Non-Amenable Finitely Presented Torsion-By-Cyclic Groups

NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS

by ALEXANDER YU. OL’SHANSKII and MARK V. S A P I R

ABSTRACT

We construct a finitely presented non-amenable group without free non-cyclic subgroups thus providing a finitely presented counterexample to von Neumann’s problem. Our group is an extension of a group of finite exponent n 1 by a cyclic group, so it satisfies the identity [x, y]n = 1.

CONTENTS 1 Introduction...... 44 1.1 Shorthistoryoftheproblem...... 44 1.2 CongruenceextensionembeddingsofthefreeBurnsidegroups...... 46 1.3 TheschemeoftheproofofTheorem1.3andtheplanofthepaper...... 50 2 Mapsanddiagrams...... 53 2.1 Graphsandmaps...... 53 2.2 Bands...... 54 2.3 Bondsandcontiguitysubmaps...... 56 2.4 ConditionA...... 59 3 Burnsidequotients ...... 64 3.1 Axioms...... 64 3.2 Corollariesof(Z1),(Z2),(Z3) ...... 67 3.3 Theconstructionofagradedpresentation...... 74 4 SubgroupsofthefreeBurnsidegroups...... 93 4.1 Setsofwordswithsmallcancellation...... 93 4.2 Subgroups of B(m, n) satisfyingthecongruenceextensionproperty...... 95 4.3 Burnsidegroupsandfreeproducts...... 104 5 Thepresentationofthegroup ...... 107 5.1 S-machinesandtheirinterpretation...... 107 5.2 The S-machine M ...... 111 5.3 Thepresentation...... 114 5.4 The subgroup A of H is a torsion group of exponent n ...... 115 6PropertiesoftheS-machine M ...... 116 6.1 The inverse semigroup P(M) ...... 116 6.2 Stabilizers in P(M) ofsubwordsofthehub...... 123 6.3 Stabilizersofarbitraryadmissiblewords...... 126 7 Bands,annuli,andtrapezia...... 142 7.1 Deﬁnitionsandbasicfacts...... 142 7.2 Forbiddenannuli...... 143 7.3 Trapezia...... 144 8ThegroupHra ...... 147 9ThegroupHkra ...... 151 9.1 Getting rid of Burnside ra-cells...... 151 9.2 Conditions(Z1),(Z2),(Z3)...... 159 10ProofofTheorem1.3...... 165 References...... 166 Subjectindex...... 167

Both authors were supported in part by the NSF grant DMS 0072307. In addition, the research of the ﬁrst author was supported in part by the Russian Fund for Basic Research 99-01-00894 and by the INTAS grant, the research of the second author was supported in part by the NSF grant DMS 9978802. 44 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

1. Introduction

1.1. Short history of the problem Hausdorff [16] proved in 1914 that one can subdivide the 2-sphere minus a countable set of points into 3 parts A, B, C, such that each of these three parts can be obtained from each of the other two parts by a rotation, and the union of two of these parts can be obtained by rotating the third part. This implied that one cannot define a finitely additive measure on the 2-sphere which is invariant under the group SO(3). In 1924 Banach and Tarski [4] generalized Hausdorff ’s result by proving, in particular, that in the Euclidean 3-space, every two bounded sets A, B with non-empty = n = n interiors can be decomposed A i=1 Ai, B i=1 Bi such that Ai can be rotated to Bi, i = 1,...,n (the so called Banach-Tarski paradox). Von Neumann [25] was first who noticed that the cause of the Banach-Tarski paradox is not the geometry of Eu- clidean 3-space but an algebraic property of the group SO(3). He introduced the concept of an amenable group (he called such groups “measurable”) as a group G which has a left invariant finitely additive measure1 µ, µ(G) = 1, noticed that if a group is amenable then any set it acts upon freely also has an invariant measure and proved that a group is not amenable provided it contains a free non-abelian subgroup. He also showed that groups like PSL(2, Z), SL(2, Z) contain free non-abelian subgroups. So analogs of Banach-Tarski paradox can be found in the Euclidean 2-space and even on the real line. Von Neumann showed that the class of amenable groups contains abelian groups, finite groups and is closed under taking subgroups, extensions, and infinite unions of increasing sequences of groups. Day and Specht showed that this class is closed under homomorphic images. The class of groups without free non-abelian subgroups is also closed under these operations and contains abelian and finite groups. The problem of existence of non-amenable groups without non-abelian free subgroups probably goes back to von Neumann and became known as the “von Neu- mann problem” in the fifties. As far as we know, the first paper where this problem was formulated was the paper by Day [10]. It is also mentioned in the monograph by Greenleaf [13] based on his lectures given in Berkeley in 1967. Tits [38] proved that every non-amenable matrix group over a field of characteristic 0 contains a non- abelian free subgroup. In particular every semisimple Lie group over a field of characteristic 0 contains such a subgroup (see also Vershik [39]). First counterexamples to the von Neumann problem were constructed by Ol’shanskii [29]. He proved that the groups with all proper subgroups cyclic constructed by him, both torsion-free [27] and torsion [28] (the so called “Tarski monsters”), are not amenable. Later Adian [1] showed that the free Burnside group of odd

1 Later amenable groups were characterized in many different ways. There exist a geometric characterization by Følner [11], a characterization by Kesten [19] in terms of random walks, and a combinatorial characterization by Grigorchuk [14] among others. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 45 exponent n ≥ 665 with at least two generators (that is the group B(m, n) given by m generators and all relations of the form vn = 1 where v is a word in generators) is not amenable. It is interesting that the possibility of using torsion groups and, in particular, free Burnside groups as potential counterexamples to von Neumann’s problem was mentioned by Day [10], conjectured by Kesten [20], and by B.H. Neumann’s in his review of Specht’s paper [37] in Zentralblatt. Both Ol’shanskii’s and Adian’s examples are not finitely presented: in the mod- ern terminology these groups are inductive limits of word hyperbolic groups, but they are not hyperbolic themselves. Since many mathematicians (especially topologists) are mostly interested in groups acting “nicely” on manifolds, it is natural to ask if there exists a finitely presented non-amenable group without non-abelian free subgroups. This question was explicitly formulated, for example, by Grigorchuk in [22] and by Co- hen in [9]. This question is one of a series of similar questions about finding finitely presented “monsters”, i.e. groups with unusual properties. Probably the most famous problem in that series is the problem about finding a finitely presented infinite torsion group. Other similar problems ask for finitely presented divisible group (group where every element has roots of every degree), finitely presented Tarski monster, etc. In each case a finitely generated example can be constructed as a limit of hyperbolic groups (see [30]), and there is no hope to construct finitely presented examples as such limits. One difficulty in constructing a finitely presented non-amenable group without free non-abelian subgroups is that there are “very few” known finitely presented groups without free non-abelian subgroups. Most non-trivial examples are solvable or “almost” solvable (see [21]), and so they are amenable. The only known examples of finitely presented groups without free non-abelian subgroups for which the problem of amenability is non-trivial, are R. Thompson’s group F and its close “relatives”. The fact that F does not contain free non-abelian subgroups was proved by Brin and Squier in [6]. A conjecture that F is not amenable was formulated first by Geoghegan [12]. A considerable amount of work has been done to prove this conjecture (see [7]) butitisstillopen. One approach to constructing a finitely presented counterexample to the von Neumann problem would be in using the Higman embedding theorem which states that every recursively presented group can be embedded into a finitely presented group. So one can take a known finitely generated non-amenable group without non-abelian free subgroups and embed it into a finitely presented group. Of course, the resulting group will be non-amenable since the class of amenable groups is closed under taking subgroups. Unfortunately all known constructions of Higman embeddings (see, for example, [34]) use amalgamated products and non-ascending HNN extensions, which immediately leads to non-abelian free subgroups. Nevertheless Higman-like embeddings play an important role in our construction. Ourmainresultisthefollowing. 46 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Theorem 1.1. — For every sufficiently large odd n, there exists a finitely presented group G which satisfies the following conditions. 1. G is an ascending HNN extension of a finitely generated infinite group of exponent n. 2. G is an extension of a non-locally finite group of exponent n by an infinite cyclic group. 3. G contains a subgroup isomorphic to a free Burnside group of exponent n with 2 generators. 4. G is a non-amenable finitely presented group without free non-cyclic subgroups.

Notice that part 1 of Theorem 1.1 immediately implies part 2. By a theorem of Adian [1], part 3 implies that G is not amenable. Thus parts 2 and 3 imply part 4. Note that the first example of a finitely presented group which is a cyclic extension of an infinite torsion group was constructed by Grigorchuk [15]. But the torsion subgroup in Grigorchuk’s group does not have a bounded exponent and his group is amenable (it was the first example of a finitely presented amenable but not elementary amenable group). In the two subsequent subsections, we present the main ideas of our construction (we simplify the notation for the sake of clarity).

1.2. Congruence extension embeddings of the free Burnside groups We first formulate two theorems on embeddings of the free Burnside group B(m, n) of sufficiently large2 odd exponent n with m>1generators, and deduce Theo- rem 1.1 from Theorems 1.2 and 1.3 which, in turn, will be proved in Sections 2–10. We say that a subgroup H of a group G satisfies the Congruence Extension Property (CEP) if any homomorphism H → H1 of H onto a group H1 extends to a homomorphism G → G1 of G onto some group G1 containing H1 as a subgroup. Equiv- alently, for any normal subgroup L of H, there exists a normal subgroup M of G such that H ∩ M = L. There is another obvious reformulation, which is more convenient when dealing with defining relations. Namely, the subgroup H satisfies CEP if, for any subset S ⊆ H,wehaveH ∩ SG = SH (where the normal closure of a subset T in a group K is denoted as TK). We will write H ≤CEP G if the subgroup H of a group G satisfies CEP. We also say that H is CEP-embedded in G,orH is a CEP-subgroup of G. Clearly every retract H of any group G satisfies CEP. It is also clear that if K is a CEP-subgroup of H and H is a CEP-subgroup of G then K is a CEP-subgroup of G.

2 The number n in the paper is chosen after one chooses several auxiliary parameters α > β > γ..., related by a system of inequalities. We ﬁrst choose α,thenβ, γ, etc. Although the values of parameters and n can be calcu- lated precisely (for example n>1010), we are not doing it here because the consistency of the system of inequalities is obvious. Indeed, each inequality has the form f (x, y,...) >0where x

Butthereexistlessobviousexamples.Forinstance,thefreegroupF∞ of inﬁnite rank can be CEP-embedded into a 2-generated free group F2.Moreover,F∞ can be CEP-embedded into any non-elementary hyperbolic group [31]. This fact was used in [31] to prove that every non-elementary hyperbolic group is SQ-universal.

Theorem 1.2. — For any sufﬁciently large odd n, there exists a natural number3 s = s(n) such that the free Burnside group B(∞, n) of inﬁnite countable rank is CEP-embedded into the free Burnside group B(s, n) of rank s.4

In the next theorem, H n denotes the subgroup of a group H generated by all n-th powers hn of elements h ∈ H .

Theorem 1.3. —Letn be a sufficiently large odd number. Then for every positive integer s, there is a finitely presented group H containing subgroup B(s, n), such that (1) H n ∩ B(s, n) = 1, so there exists a canonical embedding of B(s, n) into H /H n. (2) the embedding of B(s, n) into H /H n from (1) satisfies CEP.

To derive here Theorem 1.1 from Theorems 1.2 and 1.3, we assume that n is a sufﬁciently large odd integer and s is the number given by Theorem 1.2. Then let H be a group provided by Theorem 1.3. It has a ﬁnite presentation c1, ..., cm |R. Since retracts are CEP-subgroups, we can chose a subgroup chain

(1) B(m, n) ≤CEP B(m + 2, n) ≤CEP B(∞, n) ≤CEP Bs(s, n) ≤ H where the ﬁrst and the second embeddings are given by the embeddings of the generating sets, the third and the fourth ones are given by Theorems 1.2 and 1.3, respectively. Since CEP is a transitive property, we obtain from (1) and Theorem 1.3 the chain

(2) B(m, n) ≤CEP B(m + 2, n) ≤CEP H (n). The factor-group H (n) = H /H n admits presentation

(3) H (n) =c1, ..., cm |R ∪ V ,

n where V is the set of all n-th powers v of group words v = v(c1,...,cm). Let b1,...,bm and b1, ..., bm, bm+1, bm+2 be free generating sets of the subgroup B(m, n) and B(m + 2, n), respectively, in (1) and in (2). It will be convenient to denote these subgroups of H (and of H (n))byBb(m, n) and by Bb(m + 2, n), respectively.

3 By using our proof of Theorem 1.2 and some additional arguments, D. Sonkin recently showed that one can set s = 2. 4 Theorem 1.2 is of independent interest because it immediately implies, in particular, that every countable group of exponent n is embedable into a ﬁnitely generated group of exponent n. This was ﬁrst proved by Obraztsov (see [30]) 48 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

There are words w1(c1,...,cm), ..., wm(c1, ..., cm) such that

b1 = w1(c1, ..., cm), ..., bm = wm(c1, ..., cm) in H .Foreveryr(c1,...,cm) ∈ R,wedeﬁnederived words

r (c1, ..., cm) ≡ r(w1(c1, ..., cm), ..., wm(c1, ..., cm)), (4) r (c1, ..., cm) ≡ r (w1(c1, ..., cm), ..., wm(c1,...,cm)), ...

(the sign “≡” means the letter-for-letter equality) and deﬁne R as the set of all r , r , ..., r(i), ... for every r ∈ R. We consider the group obtained from (3) by the following formula

¯ H (n) (5) H = H (n)/(R ) =c1,...,cm |R ∪ V ∪ R . ∼ Since B(m, n) = c1, ..., cm |V ,wehavefrom(5)

¯ ∼ Bc(m,n) (6) H = Bc(m, n)/(R ∪ R ) , where Bc(m, n) is the free Burnside group with m free generators c1, ..., cm.LetRb R R R ,..., and b consist of copies of words from and written in letters b1 bm.Then from (6), we immediately deduce

( , ) ( , ) H¯ =∼ ( , )/(R ∪ R )Bc m n =∼ ( , )/(R ∪ R )Bb m n . (7) Bc m n Bb m n b b

Since the embedding of Bb(m, n) into H (n) is an CEP-embedding (see (2)), the right-hand side of (7) is isomorphic to the group

¯ = ( , )/( ( , ) ∩ (R ∪ R )H (n)). B Bb m n Bb m n b b

However, by the deﬁnition (4), r(b1, ..., bm)=r (c1, ..., cm), r (b1, ..., bm)=r (c1,...,cm), ... H ( ), (R ∪R )H (n) = (R )H (n). ¯ = ( , )/( ( , )∩(R )H (n)) in n and so b b Hence B Bb m n Bb m n , ¯ ¯ i.e. B is the canonical image of the subgroup Bb(m, n) ≤ H (n) in H . Thus, the left- ¯ ¯ ¯ hand side H of (7) is isomorphic to the image B of Bb(m, n) in H under the mapping ci → bi, i = 1, ..., m. (We preserve notations c1, ..., cm, b1, ..., bm for the images of the generators in H¯ .) The isomorphism of H¯ with its subgroup B¯ allows us to introduce the ascending HNN-extension

¯ −1 G =H , t |tcit = bi, i = 1, ..., m=c1, ..., cm |R ∪ V ∪ R ∪ U , NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 49

U ={ −1 −1, = , ..., } 5 U where tcit wi i 1 m . However, relators from and formulas (4) imply (i) −1 (i+1) (0) (1) that tr (c1, ..., cm)t = r (c1, ..., cm) for i ≥ 0 where r = r, r = r ,... for r ∈ R. Hence G =c1, ..., cm |R ∪ V ∪ U . Next we observe that for every word v in letters b1,...,bm,therelation n v(b1, ..., bm) = 1 follows from R because Bb(m, n) is a subgroup of H . Therefore n −1 n v(c1,...,cm) = t v(b1, ..., bm) t = 1 follows from R ∪ U .Hence

(8) G =c1, ..., cm |R ∪ U , and so G is a ﬁnitely presented group. The group G is an ascending HNN-extension of the group H¯,whichisofex- ponent n being a homomorphic image of the group H (n).

Finally, we notice that the canonical image B of the subgroup Bb(m + 2, n) ≤ H (n) in H¯ is isomorphic to

H ( ) n Bb(m+2,n) Bb(m + 2, n)/(R ∩ Bb(m + 2, n)) = Bb(m + 2, n)/(R ) because of the CEP-embedding Bb(m + 2, n) ≤CEP H (see (2)). The group Bb(m+2,n) Bb(m + 2, n)/(R ) can be homomorphically mapped onto the free Burnside group B(2, n) because all words from R are written in the ﬁrst m of the m + 2 generators of Bb(m + 2, n). Therefore B has a retract which is isomorphic to B(2, n).(The retraction B(2, n) → B does exist since the group B(2, n) is free in the variety of periodic group deﬁned by the identity xn = 1.) Thus, statements 1 and 3 of Theorem 1.1 (and so statements 2 and 4, as we noticed earlier) follow from Theorems 1.2 and 1.3.

Remark 1.4. — The embedding of B(s, n) into the group H in Theorem 1.3 and the set of deﬁning relation R of H will be constructed explicitly. Since the func- tion f in Lemma 4.1 and the number of generators in Lemma 4.2 are obtained explicitly, the CEP-embedding for Theorem 1.2 is constructive too. Therefore one can also writedownanexplicitpresentation(8)ofthegroupG in Theorem 1.1. But an easier, clear and explicit construction of a ﬁnite set of the subwords w1, ...wm of relators U from (8) is explained in Remark 4.3 which goes back to [35].

5 E. Rips has been publicizing a simpler idea of constructing G for several years. He suggested to consider a finitely presented group H =c1,...,cm |R, containing B(2, n), subject to relations from R and additional relations of the form ci = ui, i = 1,...,m,wherethewordsui(c1,...,cm) represent “long and complicated” elements in B(2, n). The resulting group is clearly finitely presented and torsion since the images of H and B(2, n) coincide in it. So if it were infinite, it would be an example of an infinite finitely presented torsion group (a solution of an outstanding group theory problem). If such an example were to exist, it would most certainly be non-amenable. Un- fortunately in all known cases the resulting group is obviously finite, and Rips’ idea has not been implemented yet. Our group (8) is not torsion but it is torsion-by-cyclic, and it satisfies the identity [x, y]n = 1. 50 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

1.3. The scheme of the proof of Theorem 1.3 and the plan of the paper

To prove Theorem 1.3, we have to construct special embeddings of given free Burnside group B(m, n) generated by a set B of cardinality m into a ﬁnitely presented group H such that H n ∩ B(m, n) = 1, and the canonical image of B(m, n) in H (n) = H /H n enjoys CEP. The embedding in H is obtained in a similar way as in our papers [33], [34] but we need a more complicated S-machine than in [33] (S-machines were introduced by Sapir in [36]), and now the proof requires a detailed analysis of the construction of the group H (n).

This group is a factor-group of the group H generated by C ={c1, ..., cm} subject to the relations R of H over the normal subgroup generated by Burnside relations. In other words, H (n) is the Burnside factor of H . Burnside factors of free groups and free products have been studied ﬁrst by Adian and Novikov in [26], [3]. Geomet- ric approach based on the notion of A-map was employed in the study of Burnside factors of these and more complicated groups in [30]. Papers [31], [18] extends this approach to Burnside factors of hyperbolic groups. The main problem we face in this paper is that H is “very” non-hyperbolic. In particular, the set of relations R (denoted below by Z(M, Λ)) contains many commutativity relations, so H contains non-cyclic torsion-free abelian subgroups which cannot happen in a hyperbolic group. Nevertheless (and it is one of the main ideas of the paper) one can make the Cayley graph of H look hyperbolic if one divides the generators from C into two sets and consider letters from one set as zero letters, and the corresponding edges of the Cayley graph as edges of length 0. Thus the length of a path in the Cayley graph or in a van Kampen diagram over the presentation of H is the number of non-zero edges of the path. More precisely, the group H is similar to the group G(M) of [36], [34], [33]. As we mentioned above it corresponds to an S-machine M.ThesetC consists of tape letters (the set A), state letters (the set K) and command letters (the set R). Recall that unlike an ordinary Turing machine, an S-machine works with elements of a group, not elements of the free semigroup. It turns out that the most productive point of view is to consider an S-machine as an inverse semigroup of partial transformations of a set of states which are special ∗ ∪ (admissible) words of a certain group Hka generated by K A. The generators of the semigroup are the S-rules.ThemachineM is the set of the S-rules. Every computation of the machine corresponds to a word over M which is called the history of computation, i.e. the string of commands used in the computation. With every computation h applied to an admissible element W, one associates a van Kampen diagram T(W, h) ∗ (called a trapezium) over the presentation of Hka (see Figure 1; a more precise picture of a trapezium is Figure 14 in Section 7.3). The ﬁrst and the last words of the computation are written on the bases the trapezium, copies of the history of the computation are written on the vertical sides. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 51

The horizontal strips (bands) of cells correspond to applications of individual rules in the computation. The trapezia T(W, h) play central role in our study of the Burnside factor H (n) of H . As in [30], the main idea is to construct a graded presentation of H (n) where longer relations have higher ranks and such that every van Kampen diagram over the presentation of H (n) has the so called property A from [30]. In all diagrams over the graded presentation of H (n), cells corresponding to the relations from R are considered as 0-cells or cells of rank 1/2, and cells corresponding to Burnside relations from the graded presentation are considered as cells of ranks 1, 2,.... So in these van Kampen

diagrams “big” Burnside cells are surrounded by “invisible” 0-cells and “small” cells.

ast word · L W h

History History h h

ord Starting w W

FIG. 1.

The main part of property A is the property that if a diagram over H (n) contains two Burnside cells Π1, Π2 connected by a rectangular contiguity subdiagram Γ of rank 0 where the sides contained in the contours of the two Burnside cells are “long enough” then these two cells cancel, that is the union of Γ, Π, Π can be replaced by a smaller subdiagram. This is a “graded substitute” to the classic property of small cancellation diagrams (where contiguity subdiagrams contain no cells). Roughly speaking nontrivial contiguity subdiagrams of rank 0 turn out to be trapezia of the form T(W, h) (after we clean them of Burnside 0-cells), so properties of contiguity subdiagrams can be translated into properties of the machine M,and the inverse semigroup described above. As an intermediate step in studying the group H (n), we construct a graded presentation of the Burnside factor of the subgroup of H generated by R ∪ A.Toavoid repeating the same arguments twice, for H and for the subgroup, we formulate certain key properties (Z1), (Z2), (Z3) of a presentation of a group with a separation of generators into zero and non-zero generators, so that there exists a graded presentation of the Burnside factor of the group which satisﬁes property A (Section 3). 52 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

In order to roughly explain these conditions, consider the following example. Let P = FA×FB be the direct product of two free groups of rank m. Then the Burnside factor of P is simply B(m, n)×B(m, n). Nevertheless the theory of [30] cannot be formally applied to P. Indeed, there are arbitrarily thick rectangles corresponding to relations u−1v−1uv = 1 in the Cayley graph of P so diagrams over P arenotA-mapsintheter- minology of [30] (i.e. they do not look like hyperbolic spaces). But one can obtain the Burnside factor of P in two steps. First we factorize FA to obtain Q = B(m, n) × FB. After that we consider all edges labeled by letters from A in the Cayley graph of Q as edges of length 0. As a result the Cayley graph of Q becomes a hyperbolic space. This allows us to apply the theory of A-maps from [30] to obtain the Burnside factor of Q . The real reason for the theory from [30] to work in Q is that Q satisfies our conditions (Z1), (Z2), (Z3). But the class of groups satisfying these conditions is much bigger and includes groups corresponding to S-machines considered in this paper. In particular (Z3) holds in Q because all 0-letters centralize FB. This does not happen in more complicated situations. But we associate with every cyclically reduced non-0-element w a “personal” subgroup 0(w) consisting of 0-elements which is normalized by w. The plan of the paper is the following. First in Section 2.2 we discuss some basic facts about van Kampen diagrams and the main tools of studying them. In Section 2, we repeat the main definitions and the main results of the theory of A-maps from [30, Chapter 5]. We also give some modifications of statements from [30] that we need later. Then in Section 3, we introduce properties (Z1), (Z2), (Z3) and study Burnside factors. Although we follow the general scheme of [30], we encounter new significant difficulties. One of the main difficulties is that non-zero elements can be conjugates of zero elements. As a bi-product of this study, we show that under certain mild conditions, one can one can construct analogs of HNN extensions in the class of Burnside groups of sufficiently large odd exponents (see Corollary 3.40). In Section 4, we prove some properties of the free Burnside group, in particular the existence of a countably generated free Burnside subgroup with congruence extension property (Theorem 4.4). Roughly speaking, we obtain the required embedding of the group B(∞, n) into B(s, n) if we choose the images of free generators of B(∞, n) as aperiodic words A1, A2, ... in generators of B(s, n) satisfying a strong small cancellation condition. Here we start the inductive proof with the factor-group of the absolutely free group Fs by arbitrary relations of form r(A1, A2, ...). (This first step was described in [31].) Then in Section 5, we define the presentation of our group H by listing the rules of an S-machine and showing how to convert the S-machine into a group. In Section 6 we conduct a detailed study of the inverse semigroup corresponding to our S-machine. Two important properties reflect the flavor of that section and can be singled out. Lemma 6.5 (iv) says that if there is a computation starting with an element W, with history of the form h2 (thatisifW is in the domain of h2), then there exists a computation with history hs for every integer s. In other words, if a se- NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 53 quence of rules can be applied twice in a row, it can be applied any number of times. That lemma is used in the geometric part of the paper when we are cleaning contiguity subdiagrams of Burnside 0-cells (see Lemma 9.1). Proposition 6.17 provides an important property of stabilizers of elements in the inverse semigroup generated by the S-rules. It says, basically, that if W is stabilized by h, ghg−1 and g−1hg then it is stabilized by gshg−s for any integer s. This proposition is used when we consider long and narrow contiguity subdiagrams (see Lemma 9.9). Another very important feature of our S-machine is that two words h1 and h2 in the alphabet M define the same partial transformation on the intersection of their domains provided h1 and h2 are equal modulo Burnside relations (see Lemma 6.5 (ii)). Thus the work of the S-machine is “compatible” with Burnside relations. In the remaining sections of the paper, we show that properties (Z1), (Z2), (Z3) are satisfied by the presentation of the group Hkra corresponding to our S-machine. The proof ends in Section 10.

Acknowledgment The authors are grateful to Eliyahu Rips for inspiring discussions of topics related to this paper. The authors are also grateful to Goulnara Arjantseva who read the original version of the paper and sent us numerous comments about it. This allowed us to signiﬁcantly improve the exposition.

2. Maps and diagrams Planar maps and van Kampen diagrams are standard tools to study complicated groups (see, for example, [23] and [30]). Here we present the main concepts related to maps and diagrams. A van Kampen diagram is a labeled map, so we start by dis- cussing maps.

2.1. Graphs and maps We are using the standard definition of a graph. In particular, every edge has a direction, and every edge has an inverse edge (having the opposite direction). For every edge e, e− and e+ are the beginning and the end vertices of the edge. A map is simply a finite plane graph drawn on a disc or an annulus which sub- divides this surface into polygonal cells. Edges do not have labels, so no group presen- tations are involved in studying maps. On the other hand the theory of maps helps find the structure of van Kampen diagrams because every diagram becomes a map after we remove all the labels. 54 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Let us recall the necessary definitions from [30]. A map is called graded if every cell Π is assigned a non-negative number, r(Π), its rank.Amap∆ is called a map of rank i if all its cells have ranks ≤ i.Everygraded map ∆ has a type τ(∆) which is the vector whose first coordinate is the maximal rank of a cell in the map, say, r, the second coordinate is the number of cells of rank r in the map, the third coordinate is the number of cells of rank r − 1 and so on. We compare types lexicographically. Let ∆ be a graded map on a surface X. The cells of rank 0 in ∆ are called 0-cells.Someoftheedgesin∆ are called 0-edges. Other edges and cells will be called positive.Wedefinethelength of a path p in a map as the number of positive edges in it. The length of a path p is denoted by |p|,inparticular,|∂(Π)| is the perimeter of acellΠ. We assume that (M1) the inverse edge of a 0-edge is also a 0-edge, (M2) either all edges in a 0-cell are 0-edges or exactly two of these edges are positive, (M3) for every cell Π of rank r(Π) >0the length |∂(Π)| of its contour is positive.

2.2. Bands

A van Kampen diagram over a presentation X | R is a planar map with edges labeled by elements from X ±1, such that the label of the contour of each cell belongs to R up to taking cyclic shifts and inverses [30]. In §11 of [30], 0-edges of van Kampen diagrams have label 1, but in this paper 0-edges will be labeled also by the so called 0-letters. All diagrams which we shall consider in this paper, when considered as planar maps, will obviously satisfy conditions (M1), (M2), (M3) from Section 2.1. ByvanKampenLemma,awordw over X is equal to 1 modulo R if and only if there exists a van Kampen disc diagram over R with boundary label w. Similarly (see [23]) two words w1 and w2 are conjugate modulo R if and only if there exists an annular diagram with the internal counterclockwise contour labeled by w1 and external counterclockwise contour labeled by w2.Theremaybemanydiagramswiththesame boundary label. Sometimes we can reduce the number of cells in a diagram or ranks of cells by replacing a 2-cell subdiagram by another subdiagram with the same boundary label. The simplest such situation is when there are two cells in a diagram which have a common edge and are mirror images of each other. In this case one can cancel these two cells (see [23]). In order strictly define this cancellation (to preserve the topological type of the diagram), one needs to use the so called 0-refinement. The 0-refinement consists of adding 0-cells with (freely trivial) boundary labels of the form 1aa−1 or 111 (see [30], §11. 5). NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 55

As in our previous papers [36], [5] and [32], one of the main tools to study van Kampen diagrams are bands and annuli. Let S be a subset of X .AnS-band B is a sequence of cells π1, ..., πn,forsome n ≥ 0 in a van Kampen diagram such that • Each two consecutive cells in this sequence have a common edge labeled by a letter from S. • Each cell πi, i = 1,...,n has exactly two S-edges (i.e. edges labeled by a letter from S). • If n = 0, then the boundary of S has form ee−1 for an S-edge e.

Figure 2 illustrates this concept. In this figure edges e, e1, ..., en−1, f are S-edges, the lines l(πi, ei), l(πi, ei−1) connect fixed points in the cells with fixed points of the corresponding edges.

q ... 2 π1 e1 π2 e2 S en−1 πn e f q1 ... l(π1,e) l(π1,e1) l(π2,e1) l(π2,e2) l(πn−1,en−1) l(πn,en−1) l(πn,f)

FIG. 2.

Thebrokenlineformedbythelinesl(πi, ei), l(πi, ei−1) connecting points inside neighboring cells is called the connecting line6 of the band B.TheS-edges e and f are called the start and end edges of the band. The counterclockwise boundary of the subdiagram formed by the cells π1, ..., πn B −1 −1 B B of has the form e q1 fq2 .Wecallq1 the bottom of and q2 the top of , denoted bot(B) and top(B). Abandπ1,...,πt is called reduced if πi+1 is not a mirror image of πi, i = 1, ..., t−1 (otherwise cells πi and πi+1 cancel and there exists a diagram with the same boundary label as ∪iπi ad containing fewer cells). Similarly, we deﬁne reduce annuli. We say that two bands cross if their connecting lines cross. We say that a band is an annulus if its connecting line is a closed curve. In this case the ﬁrst and the last cellsofthebandcoincide(seeFigure3a). If a band B connects two edges on the boundary of a van Kampen disc diagram ∆ then the connecting line of B divides the region bounded by ∂(∆) into two parts. Let ∆1 and ∆2 be the maximal subdiagrams of ∆ contained in these parts. Then we say that the (connecting line of the) band B divides ∆ into two subdiagrams ∆1

6 In [36], [5] this line was called the median of the band, but in [30], it is called the connecting line. 56 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

S πn π3 π1 πn−1 π2 π = π γ γm

1 n π 1 T π

a b

FIG. 3.

and ∆2.NoticethatB does not belong to either ∆1 or ∆2 and ∆ = ∆1 ∪ ∆2 ∪ B. Similarly if ∆ is an annular diagram and B is an annulus surrounding the hole of ∆ then (the connecting line of) B divides ∆ into two annular subdiagrams. The maximal subdiagram bounded by the connecting line of an annulus in a disc diagram is called the inside diagram of this annulus.

Let S and T be two disjoint subsets of X ,let(π, π1, ..., πn, π )beanS-band and let (π, γ1, ..., γm, π )isaT-band. Suppose that: • the connecting lines of these bands form a simple closed curve, • on the boundary of π and on the boundary of π the pairs of S-edges separate the pairs of T-edges, • the start and end edges of these bands are not contained in the region bounded by the connecting lines of the bands. Then we say that these bands form an (S, T)-annulus and the closed curve formed by the parts of connecting lines of these bands is the connecting line of this annulus (see Figure 3b). For every (S, T)-annulus we deﬁne the inside subdiagram diagram of the annulus as the maximal subdiagram bounded by the connecting line of the annulus.

2.3. Bonds and contiguity submaps Now let us introduce the main concepts from Chapter 5 of [30]. Consider amap∆.Apathin∆ is called reduced if it does not contain subpaths of length 2 which are homotopic to paths of length 0 by homotopies involving only 0-cells. We shall usually suppose that the contour of ∆ is subdivided into several subpaths called sections. We usually assume that sections are reduced paths. By property M2 if a 0-cell has a positive edge, it has exactly two positive edges. Thus we can consider bands of 0-cells having positive edges. The start and end edges of such bands are called adjacent. Let e and f be adjacent edges of ∆ belonging to two positive cells Π1 and Π2 or to sections of the contour of ∆. Then there exists a band of 0-cells connecting these edges. The union of the cells of this band is called a 0-bond between the cells Π1 and NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 57

Π2 (or between a cell and a section of the contour of ∆, or between two sections of the contour). The connecting line of the band is called the connecting line of the bond. The contour of the 0-bond has the form p−1es−1f −1 where p and s are paths of length 0 because every 0-cell has at most two positive edges.

Now suppose we have chosen two pairs {e1, f1} and {e2, f2} of adjacent edges such , Π −1, −1 Π that e1 e2 belong to the contour of a cell 1 and f1 f2 to the contour of a cell 2. Then we have two bonds E1 and E2.IfE1 = E2 (that is e1 = e2, f1 = f2)thenlet Γ = = −1 −1 E1.NowletE1 E2 and z1e1w1 f1 and z2e2w2 f2 be the contours of these bonds. Further let y1 and y2 be subpaths of the contours of Π1 and Π2 (of Π1 and q, or q1 and q2,whereq, q1 and q2 are some segments of the boundary cotours) where y1 −1 −1 (or y2)hastheforme1pe2 or e2pe1 (or ( f1uf2) or ( f2uf1) ). If z1 y1w2 y2 (or z2 y1w1 y2) is a contour of a disc submap Γ which does not contain Π1 or Π2,thenΓ is called a 0-contiguity submap of Π1 to Π2 (or Π1 to q or of q1 to q2). The contour of Γ is naturally subdivided into four parts. The paths y1 and y2 are called the contiguity arcs.We write y1 = Γ Π1, y2 = Γ Π2 or y2 = Γ q. The other two paths are called the side arcs of the contiguity submap.

The ratio | y1|/|∂(Π1)| (or | y2|/|∂(Π2)| is called the degree of contiguity of the cell Π1 to the cell Π2 or to q (or of Π2 to Π1). We denote the degree of contiguity of Π1 to Π2 (or Π1 to q)by(Π1, Γ, Π2) (or (Π1, Γ, q)). Notice that this deﬁnition is not symmetric and when Π1 = Π2 = Π,forexample,then(Π, Γ, Π) is a pair of numbers.

A connecting line of Γ is chosen as the connecting line of one of the bonds E1 or E2. We say that two contiguity submaps are disjoint if none of them has a common point with the interior of the other one, and their contiguity arcs do not have common edges. Now we are going to define k-bonds and k-contiguity submaps for k>0.In these definitions we need a fixed number ε, 0<ε <1. Let k>0and suppose that we have defined the concepts of j-bond and j-contiguity submap for all j

1. r(π) = k, r(Π1) >k, r(Π2) >k, 2. there are disjoint submaps Γ1 and Γ2 of j1-contiguity of π to Π1 and of j2- contiguity of π to Π2, respectively, with j1

Then there is a minimal submap E in ∆ containing π, Γ1, Γ2. This submap is called a k-bond between Π1 and Π2 deﬁned by the contiguity submaps Γ1 and Γ2 with principal cell π (see Figure 4). 58 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Π1

q1 v1 t π v2

q2 Π2

FIG. 4. Π Γ Π The contiguity arc q1 of the bond E to 1 is 1 1. It will be denoted by E Π1. Similarly E Π2 is by definition the arc q2 = Γ2 Π2. The contour ∂(E) canbewrittenintheformp1q1p2q2 where p1, p2 are called the side arcs of the bond E. For each k-bond we fix a connecting line obtained by joining the terminal points of connecting lines of Γ1 and Γ2 inside π. The connecting line divides π (and the whole E) into two connected components. Bonds between a cell and a section of the contour of ∆ or between two sections of the contour are defined in a similar way.

Now let E1 be a k-bond and E2 be a j-bond between two cells Π1 and Π2, j ≤ k and either E1 = E2 or these bonds are disjoint. If E1 = E2 then Γ = E1 = E2 is called the k-contiguity submap between Π1 and Π2 determined by the bond E1 = E2.IfE1 and E2 are disjoint then the corresponding contiguity submap Γ is deﬁned as the smallest submap containing E1 and E2 and not containing Π1 and Π2 (see Figure 5).

FIG. 5. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 59

The contiguity arcs q1 and q2 of Γ are intersections of ∂(Γ) with ∂(Π1) and ∂(Π2). The contour of Γ has the form p1q1p2q2 where p1 and p2 are called the side arcs of Γ. The connecting line of Γ is one of the connecting lines of E1 or E2.Theratio|q1|/|∂(Π1)| is called the contiguity degree of Π1 to Π2 with respect to Γ and is denoted by (Π1,Γ,Π2). If Π1 = Π2 = Π then (Π, Γ, Π) is a pair of numbers. Contiguity submaps of a cell to a section of the contour and between sections of the contour are defined in a similar way. (In this paper we do not use notion “degree of contiguity of a section of a contour to” anything.) We are going to write “bonds” and “contiguity submaps” in place of “k-bonds” and “k-contiguity submaps”. Instead of writing “the contiguity submap Γ of a cell Π to ...”, we sometimes write “the Γ-contiguity of Π to ...”. The above definition involved the standard decomposition of the contour of a Γ = Γ Π = Γ Π contiguity submap into four sections p1q1p2q2 where q1 1, q2 2 (or q2 = Γ q if q is a section of ∂(∆)), we shall write p1q1p2q2 = ∂(Π1, Γ, Π2),andsoon. As in [30] (see §15. 1)wefixcertainrealnumbersι ζ ε δ γ β α between0and1where“” means “much smaller”. Here “much” means α¯ = 1 + α enough to satisfy all the inequalities in Chapters 5, 6 of [30]. We also set 2 , β¯ = 1 − β, γ¯ = 1 − γ, h = δ−1, n = ι−1.

2.4. Condition A The set of positive cells of a diagram ∆ is denoted by ∆(2) and as before the length of a path in ∆ is the number of positive edges in the path. A path p in ∆ is called geodesic if |p|≤|p | for any path p combinatorially homotopic to p. The condition A has three parts: A1 If Π is a cell of rank j>0, then |∂(Π)|≥nj. A2 Any subpath of length ≤ max( j, 2) of the contour of an arbitrary cell of rank j in ∆(2) is geodesic in ∆. A3 If π, Π ∈∆(2) and Γ is a contiguity submap of π to Π with (π, Γ, Π) ≥ ε, then |Γ Π| < (1 + γ)k where k = r(Π). A map satisfying conditions A1, A2, A3 will be called an A-map. As in [30], Section 15.2, a (cyclic) section q of a contour of a map ∆ is called a smooth section of rank k>0(we write r(q) = k)if: 1) every subpath of q of length ≤ max(k, 2) of q is geodesic in ∆; 2) for each contiguity submap Γ of a cell π to q satisfying (π, Γ, q) ≥ ε,wehave |Γ q| < (1 + γ)k. It is shown in [30], §§16, 17,thatA-maps have several “hyperbolic” properties. We summarize these properties here. (They show that A-maps of rank i are hyperbolic spaces with hyperbolic constant depending on i only.) 60 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

The next three lemmas are Lemmas 15.3, 15.4 and 15.8 in [30]. Lemma 2.1 shows that in a contiguity diagram between two cells side arcs are “much shorter” than perimeters of these cells (so the cells are close to each other).

Lemma 2.1. —Let∆ be an A-map, Γ a contiguity submap of either:

(i) a cell Π1 to a cell Π2 or 1 (ii) a cell Π1 to a boundary section q of ∆,or (iii) a boundary section q1 to a boundary section q2. −1 Let p1 and p2 be the side arcs of Γ,andc = max(|p1|, |p2|).Thenc<2ε r(Π1) < 1 ζnr(Π1) in cases (i) and (ii), and c<ζnr(Π2) in case (i). Furthermore c<ζnr(q ) in cases (ii) and (iii) provided q1 is smooth.

Lemma 2.2 says that contiguity arcs in a contiguity subdiagram have “almost the same lengths”. ψ Γ Π Π Lemma 2.2. —Let be the degree of -contiguity of a cell 1 to a cell 2 (or to a section q of a contour) in an A-map ∆,andq1 = Γ Π1, q2 are the contiguity arcs of Γ. Then

¯ −1 (β − 2ζψ )|q1| < |q2|. | | (ψ − β)|∂(Π )|, ψ ≥ ε, | | ( + β)| |. In particular, q2 > 2 1 and if also then q1 < 1 2 q2 Furthermore if q2 = Γ Π2 or q2 = Γ q and q is a smooth section, then

¯ −1 −1 |q1| > β(1 + 2ζψ ) |q2|. In particular, if ψ ≥ ε,then

|q1| > (1 − 2β)|q2|.

Lemma 2.3 shows that the degree of contiguity of one cell to another cell cannot be signiﬁcantly bigger than 1/2 and the degree of contiguity of a cell of a bigger rank to a cell of a smaller rank is very small.

Lemma 2.3. —InanarbitraryA-map∆ the degree of contiguity of an arbitrary cell π to an arbitrary cell Π or to an arbitrary smooth section q of the contour via an arbitrary contiguity submap Γ is less than α¯, and the degree of contiguity of π to Π (or to q) with r(Π) ≤ r(π) (or r(q) ≤ r(π)) is less than ε.

The next lemma is a generalization of the Greenlinger lemma from the theory of small cancellation groups [23]. It shows that any A-map contains a cell whose boundary is almost entirely contiguous to the boundary of the diagram. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 61

Lemma 2.4. —a)Let∆ be a disc A-map whose contour is subdivided into at most 4 sections q1, q2, q3, q4,andr(∆) >0. Then there exists a positive cell π and disjoint contiguity Γ ,...,Γ π 1,..., 4 submaps 1 4 of to q q (some of these submaps may be absent) such that the sum of 4 (π, Γ , i) γ¯ = − γ corresponding contiguity degrees i=1 i q is greater than 1 . b) Let ∆ be an annular A-map whose contours are subdivided into at most 4 sections q1, q2, q3, q4 regarded as cyclic or ordinary paths, and r(∆) >0.Then∆ has a positive cell π and 1 2 3 4 disjoint contiguity submaps Γ1,..., Γ4 of π to q q , q , q (up to three of these submaps may be absent) such that the sum of contiguity degrees of these contiguity submaps is greater than γ¯ .

Proof. — Both statements are immediate corollaries of Theorem 16.1 from [30].

The next lemma follows from Lemma 2.4 in the same way as Theorem 16.2 follows from Corollary 16.1 in [30].

Lemma 2.5. —If∆ is a disc or annular A-map with contour subdivided into at most 4 sections, and r(∆) >0, then there exists a positive cell Π and a contiguity submap Γ of Π to one of the sections q of the contour of ∆ such that r(Γ) = 0 and (Π, Γ, q) ≥ ε.

Here we add an additional property whose proof is contained in the proof of Theorem 16.2 in [30].

Lemma 2.6. —Let∆ be a disc or annular A-map with contour subdivided into at most 4 sections. Assume there is a cell Π in ∆ and a contiguity submap Γ of Π to one of the sections q of the boundary with (Π, Γ, q) ≥ ε. Then there is a cell π of ∆ and a contiguity submap Γ of π to q such that r(Γ ) = 0 and (π, Γ , q) ≥ ε.

The next four lemmas are Theorem 17.1, Corollary 17.1, Lemma 17.1 and Lemma 17.2 in [30]. The following lemma shows that smooth sections of contours of A-maps are almost geodesic.

Lemma 2.7. —Let∆ be a disc map with contour qt or an annular A-map with contours q and t. If q is a smooth section then β¯|q|≤|t|.

ThenextninelemmasdemonstratedifferentvariantsofthefactthatinA-maps, quasigeodesic quadrangles are thin. But we need more precise statements than in the usual theory of hyperbolic spaces. The constants appearing in these statements do not depend on the rank of diagrams while the hyperbolic constants of A-maps tends to inﬁnity with the rank.

Lemma 2.8. —IfadiscA-map ∆ contains a cell Π of positive rank, then |∂(∆)| > β¯ |∂(Π)|. 62 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Lemma 2.9. —Let∆ be a disc A-map with contour p1q1p2q2 where q1 and q2 are smooth sections such that |q1|+|q2| >0, |p1|+|p2|≤γ(|q1|+|q2|). Then there is either a 0-bond between q1 and q2 or a cell Π ∈ ∆(2) and disjoint contiguity submaps Γ1 and Γ2 of Π to q1 ¯ and q2, respectively, such that (Π, Γ1, q1) + (Π, Γ2, q2) > β = 1 − β. We also need the following version of Lemma 2.9. We say that a bond between two boundary sections q1 and q2 of a map is narrow if it is either a 0-bond or, in the inductive definition of the bond, we add that (π, Γ1, q1) + (π, Γ2, q2) >1− α/2 for the principal cell π of the bond. A contiguity submap between q1 and q2 is narrow if it is defined by narrow bonds (according to the inductive definition).

Lemma 2.10. —Let∆ be a disc A-map with contour p1q1p2q2 where q1 and q2 are 2 smooth sections such that |q1|+|q2| >0, |p1|+|p2|≤α (|q1|+|q2|). Then there is either a 0-bond between q1 and q2 or a cell Π ∈ ∆(2) and disjoint contiguity submaps Γ1 and Γ2 of Π to q1, q2 (respectively) such that (Π, Γ1, q1) + (Π, Γ2, q2) ≥ 1 − α/2. Hence Π, Γ1 and Γ2 form a narrow bond between q1 and q2. Proof. — The difference between the formulations of Lemma 2.10 and Lem- ma 17.2 of [30] is that the parameters γ and β¯ are now substituted by α2 and 1 − α/2, respectively. The proof differs from the proof of Lemma 17.2 [30] only in some insigniﬁcant details. There is no change in Case 1) of the proof of Lemma 17.2 [30]. The only alteration in Case 2) is the substitution of the ﬁrst factor γ for α2 in the left-hand side of the long inequality at the end. It is easy to see that the resulting inequality is also true by the choice of the parameters. In Case 3), we should assume now that (Π, Γ, p1) > γ¯ − (1 − α/2) = α/2 − γ ¯ (instead of (Π, Γ, p1) > γ¯ − β = β − γ in [30]). This leads to the similar substitution of the term β − γ for α/2 − γ in the right-hand side of inequality (7) of the proof. It remains to notice that the last long inequality of Case 3) is still true after the substitution of β − γ by α/2 − γ in the right-hand side, and the substitution of the factor γ by α2 in the left-hand side. The difference β−γ should be replaced by α/2−γ in the hypothesis of Case 4). This leads to the same substitution in inequality (9) of the proof of Lemma 17.2 [30]. Then again, it remains to replace the term β − γ by α/2 − γ in the right-hand side of the concluding inequality, and to replace the factor γ by α2 in the left-hand side of it.

Lemma 2.11. —Let∆ be a disc A-map with contour p1q1p2q2, where q1 and q2 are smooth sections of some ranks k and l, k ≤ l, and |p1|+|p2|≤γ(|q1|+|q2|). Then there , = is a narrow contiguity submap between q1 and q2 with contour p1q1p2q2 where q1 u q1u , = | |, | | γ −1(| |+α ), | |, | | γ −1(| |+α ). q2 v q2v and u v < p1 k u v < p2 k NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 63

Proof. — Up to the term “narrow”, the proof is the same as the proof of items 1) and 2) of Lemma 17.3 [30] (Lemma 2.12 below). In fact Lemma 2.11 follows from the proof of Lemma 17.3 [30] because α/2>β.

Lemma 2.12 (Lemma 17.3 [30]). — Let ∆ be a disc A-map with contour p1q1p2q2 where q1 and q2 are smooth sections of ranks k and l, k ≤ l, such that

(9) |p1|+|p2|≤γ(|q1|+|q2|). Then:

(1) there exist vertices o1 and o2 on q1 and q2 and a path x joining them in ∆ such that |x| < αk. (2) one can choose o1 and o2 in such a way that the lengths of the initial segments of q1 and −1 −1 , γ −1(| |+α ) q2 (or of q1 and q2)endingato1 and o2 are less than p1 k (or than −1 γ (|p2|+αk)). (3) if (9) is replaced by

|p1|, |p2| < αk,

then any vertex o of q1 (or of q2) can be joined in ∆ by a path y to a vertex o on q2 −1 (or on q1) in such a way that | y| < γ k.

Consider an A-map ∆ with contour p1q1p2q2, where q1, q2 are smooth sections and ∆ is a narrow contiguity submap between q1 and q2. We deﬁne an α-series as a maximal series of disjoint narrow bonds between q1 and q2. In particular, the maximal series contains the bonds from the deﬁnition of narrow contiguity submap. By Π0 we denote a cell of maximal perimeter among principal cells of the bonds over all α-series.

Lemma 2.13. — With the preceding notation, any vertex of q1 is at the distance at most −1 α |∂(Π0)| from the path q2.

, ..., α l l l l Proof. — Assume that E1 Es is the -series. Let x1 y1x2 y2 be the standard con- , l = , . tour of El where yj is a subpath of qj for j 1 2 By Lemma 2.1 and the deﬁnition of | l| (α + ζ)|∂(Π )|. | l | narrow bonds, xj < 2 0 By Lemmas 2.2 and 2.3, yj < ( − β)−1α¯|∂(Π )|. l 1 2 0 Therefore the distance between any vertex of some y1 and q2 −1 −1 is less than (α + 2ζ + (1 − 2β) α/¯ 2)|∂(Π0)| < |∂(Π0)|/3<α |∂(Π0)|. ∆ ,...,∆ ( l )−1 l ( l+1)−1 l Then we consider the submaps 1 s−1 with contours x2 z1 x1 z2, l = , . | l |+| l | α−2(| l |+ where zj is a subpath of qj for j 1 2 One may suppose that z1 z2 < x2 | l+1|), x1 because otherwise by Lemma 2.10 there would be an additional narrow bond in ∆l, which contradicts the deﬁnition of the α-series. (Indeed (Π, Γj , qj) ≥ 1 − α/2 − α¯ > ε for j = 1, 2 in the Lemma 2.10 by Lemma 2.3, and therefore that cell Π is 64 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

. | l |+|l+1| a principle cell of a narrow bond between q1 and q2 )Since x2 x1 < (α + ζ)|∂(Π )|, | l | α−1( + ζα−1)|∂(Π )|, 4 0 we have z1 < 1 4 0 and the distance between l | l |/ + (α + ζ)|∂(Π )| α−1|∂(Π )|, any vertex of z1 and q2 is less than z1 2 2 0 < 0 and the lemma is proved.

Lemma 2.14 (Lemma 17.4 [30]). — Let ∆ be a disc A-map with contour p1q1p2q2, where q1 and q2 are smooth sections of ranks k and l, k ≤ l,and|p1|, |p2| < ζnk. Then the perimeter of every cell of ∆ is less than 3γ −1ζnk < nk and r(∆) <3γ −1ζk

Lemma 2.15 (Lemma 17.5 [30]). — Let ∆ be a disc A-map with contour p1q1p2q2, where q1 and q2 are smooth sections of ranks k and l, k ≤ l,and|qj|−M ≥ 0 for j = 1, 2 and = γ −1(| |+| |+ ). ∆ M p1 p2 k Then it is possible to excise a submap from with contour p1q1p2q2 | |, | | α , | | | |− = , . where q1 (q2) is a subpath of q1 (of q2), p1 p2 < k and qj > qj M for j 1 2 ThenextlemmashowsthatannularA-maps can be cut by short paths to obtain adiscmap.

Lemma 2.16 (Lemma 17.1, [30]). — Let ∆ be an annular A-map with contours p and q such that any loop consisting of 0-edges of ∆ is 0-homotopic. Then there is a path t connecting some vertices o1 and o2 of the paths p and q respectively, such that |t| < γ(|p|+|q|).

3. Burnside quotients

3.1. Axioms Recall that in Chapter 6 of [30], it is proved that for any sufficiently large odd n, there exists a presentation of the free Burnside group B(m, n) with m ≥ 2 generators, such that every reduced diagram over this presentation is an A-map. Our goal here is to generalize the theory from [30] to Burnside factors of non- free groups in the following way. In [30], 0-edges are labeled by 1 only, and (trivial) 0-relators have the form 1·...·1 or a·1·...·1·a−1 ·1·...·1. We need to increase the collection of labels of 0-edges and the collection of 0-relations. In other words, we divide the generating set of our group into the set of 0-letters and the set of positive letters. Then in the diagrams, 0-letters label 0-edges. We shall also add defining relators of our group which have the form uavb−1 where u, v are words consisting of 0-letters only (we call such words 0-words), and a, b are a positive letters, as well as defining relations without positive letters, to the list of zero relations. The corresponding cells in diagrams will be called 0-cells. Let G be a group given by a presentation X | R.LetX be a symmetric set (closed under taking inverses). Let Y ⊆ X be a symmetric subset which we shall call the set of non-zero letters or Y-letters. All other letters in X will be called 0-letters. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 65

By the Y-length of a word A, written |A|Y or just |A|, we mean the number of occurrences of Y-letters in A.TheY-length |g|Y of an element g ∈ G is the length of the Y-shortest word representing element g. Let G(∞) be the group given by the presentation X | R inside the variety of Burnside groups of exponent n. We will choose a presentation of the group G(∞) in the class of all groups, that is we add enough relations of the form An = 1 to ensure that G(∞) has exponent n. This infinite presentation will have the form X | R(∞) =∪Si where sets Si will be disjoint, and relations from Si will be called relations of rank i. Unlike [30, Chapter 6], i runs over 0, 1/2, 1, 2, 3,....(i = 0, 1, 2,... in [30, Chapter 6].) The set R will be equal to S0 ∪ S1/2.Thedetailed description of sets Si will be given below. We shall call this presentation of G(∞) a graded presentation. Our goal is the same as in [30]: we want every diagram over R(∞) to be an A-map. We are going to prove that this goal is possible to achieve if R satisfies conditions (Z1), (Z2), (Z3) presented below. While listing these conditions, we also fix some notation and definitions.

(Z1) The set R is the union of two disjoint subsets S0 = R0 and S1/2.The group X | S0 is denoted by G(0) and is called the group of rank 0.We call relations from S0 relations of rank 0. The relations from S1/2 have rank 1/2. The group X |R is denoted by G(1/2).

(Z1.1) The set S0 consists of all relations from R which have Y-length 0 and all relations of R which have the form (up to a cyclic shift) −1 = , ∈ + ay1by2 1 where y1 y2 Y , a and b are of Y-length 0 (i.e. 0-words). n (Z1.2) The set S0 implies all Burnside relations u = 1 of Y-length 0. The subgroup of G(0) generated by all 0-letters will be called the 0-subgroup of G(0). Elements from this subgroup are called 0-elements. Elements which are not conjugates of elements from the 0-subgroup will be called essential. An essential element g from 2 G(0) is called cyclically Y-reduced if |g |Y = 2|g|Y. We shall show that g is cyclically Y- reduced if and only if it is cyclically minimal (in rank 0), i.e. it is not a conjugate of a Y-shorter element in G(0) (Lemma 3.5). Notice that condition (Z1.1) allows us to consider Y-bands in van Kampen diagrams over S0. Maximal Y-bands do not intersect.

(Z2) The relators of the set S1/2, will be called hubs. The corresponding cells in van Kampen diagrams are also called hubs. They satisfy the following properties (Z2.1) The Y-length of every hub is at least n. (Z2.2) Every hub is linear in Y, i.e. contains at most one occurrence y±1 for every letter y ∈ Y. 66 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

(Z2.3) Assume that each of words v1w1 and v2w2 is a cyclic permutation of a hub or of its inverse, and |v1|≥ε|v1w1|. Then an equality u1v1 = v2u2 for some 0-words u1, u2, implies in G(0) equality u2w1 = w2u1 (see Figure 6).

u2 u1 v1

FIG. 6.

AwordB over X is said to be cyclically minimal in rank 1/2 if it is not a conjugate in G(1/2) of a word of smaller Y-length. An element g of G(0) is called cyclically minimal in rank 1/2 if it is represented by a word which is cyclically minimal in rank 1/2. (Z3) With every essential element g ∈ G(0) we associate a subgroup 0(g) ≤ G(0),normalizedbyg.Ifg is cyclically Y-reduced, let 0(g) be the maximal subgroup consisting of 0-elements which contains g in its normalizer. By Lemmas 3.3 (2) and 3.4 (3) below arbitrary essential element g is equal to a product vuv−1 where u is cyclically Y-reduced. In this case we deﬁne 0(g) = v0(u)v−1. We shall prove in Lemma 3.7 that 0(g) is well deﬁned. (Z3.1) Assume that g is an essential cyclically minimal in rank 1/2 element of G(0) and for some x ∈ G(0),bothx and g−4xg4 are 0-elements. Then x ∈ 0(g). (Z3.2) For every essential cyclically minimal in rank 1/2 element g ∈ G(0) there exists a 0-element r such that gr−1 commutes with every element of 0(g).7

7 At the beginning of May 2002, inﬂuenced by the talk by Sergei Ivanov at the Geometric Group Theory conference in Montreal, we realized that Condition (Z3.2) can be replaced by the following weaker condition: (Z3.2’) For every essential cyclicly minimal in rank 1/2 element g ∈ G(0), the extension of 0(g) by the automorphism induced by g (acting by conjugation) satisﬁes the identity xn = 1. It is clear, that in the proof of Lemma 3.12 below (the only lemma in this paper, where (Z3.2) is used), property (Z3.2) can be replaced by (Z3.2’). NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 67

In this section, we show how the theory from [30, Chapter 6] can be adapted to diagrams over a presentation satisfying (Z1), (Z2), (Z3).

3.2. Corollaries of (Z1), (Z2), (Z3) Here we are going to deduce some useful facts about groups G(0) whose pre- sentations satisfy (Z1), (Z2), (Z3). (Some of them can be deduced from a presentation of G(0) as an HNN-extension.) The following lemma is obvious.

Lemma 3.1. —IfthesetR satisﬁes property (Z1.1) then there exists a homomorphism from G(0) to the group of integers Z which sends every Y-letter to 1 and every 0-letter to 0.

Lemma 3.2. — Suppose that R satisﬁes (Z1.1). Let ∆ be a van Kampen diagram over −1 S0 and p be a subpath of the boundary ∂(∆). Suppose that a Y-band starting on p ends on p . Then φ(p) contains a subword of Y-length 2 which is equal to a 0-word in G(0).

Proof. — Since maximal Y-bands cannot intersect, there exists a Y-band T starting on edge e and ending on edge f −1,wheree, f ∈ p, such that the shortest subpath p1 of p containing e and f has length 2.Let∆1 be the subdiagram of ∆ bounded by p1 and one of the sides q1 of T .Thewordφ(q1) is a 0-word, so φ(p1) is a subword of A of Y-length 2 which is equal to a 0-word in G(0).

AwordA which does not contain a subword of Y-length 2 that is equal to a 0- word in G(0), will be called Y-reduced.AwordA is called cyclically Y-reduced if every cyclic shift of A is Y-reduced.

Lemma 3.3. — Suppose that R satisﬁes (Z1.1). Then the following conditions hold.

(1) Let A be a word representing an element g ∈ G(0),but|A|Y > |g|Y.ThenA is not Y-reduced. (2) If A is a Y-reduced word that represents a conjugate of an essential element g, but it is not a shortest word representing a conjugate of g, then the word A has the form as a product

UVU where |U|Y >0and U U is equal to a 0-word in G(0), V is a cyclically Y-reduced word.

Proof. — (1) Let B be a Y-shortest word representing g. Then there exists a dia- −1 gram over S0 with boundary pq where φ(p) ≡ A, φ(q) ≡ B . Condition (Z1.1) imply that every Y-edge on A is the start edge of a maximal Y-band. Each of these Y-bands ends on the boundary of ∆.Since|A|Y > |B|Y,one of the Y-bands starting on p ends on p. By Lemma 3.2 there exists a subword of A of length 2 which is equal to a 0-word in G(0). 68 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

To prove part (2), one can take an annular diagram instead of a disc diagram as in part (1), and prove that a cyclic shift of A has a subword of Y-length 2 which is equal to a 0-word in G(0).HenceA ≡ UVU where |U|Y >0, U U is equal to a 0-word W in G(0).SincethewordVW represents a conjugate of g and is shorter than A, we can conclude the proof by induction on |A|Y.

Lemma 3.4. — Suppose that R satisﬁes (Z1.1). Then the following statements are true. (1) If g ∈ G(0) and A is a Y-reduced word representing g then there exists a decomposition A ≡ A1A2 such that the element represented by A2A1 is cyclically Y-reduced in G(0). m m (2) If g ∈ G(0) is cyclically Y-reduced then for every integer m, |g |Y =|m||g|Y and g is cyclically Y-reduced. (3) If g is cyclically Y-reduced and A is a Y-shortest word representing g in G(0) then A is cyclically Y-reduced.

Proof. — (1) By Lemma 3.3 (1) we can suppose that A is a Y-shortest word repre- 2 senting g.LetB be a shortest word representing g .If|B|Y = 2|A|Y then g is cyclically 2 Y-reduced. Suppose that 2|A|Y > |B|Y. By Lemma 3.3 there exists a subword C of A of Y-length 2 which is equal to a 0-word D in G(0). By Lemma 3.1 we can assume ≡ −1 ≡ −1 , ∈ + that C y1uy2 (or C y1 uy2)whereu is a 0-word and y1 y2 Y . −1 ≡ Since A is Y-reduced, one can represent A as u1 y2 A1 y1u1 where u1u2 u.Then the cyclic shift A1 y1uy2 of A can be represented by A1D which is a reduced word. Since this word is Y-shorter than A we can apply induction and conclude that there exists a cyclic shift of A1D which represents a cyclically reduced element. (2) Let A be a shortest word representing g. Suppose that for some m>0, Am is not a shortest word representing gm. Then by Lemma 3.3 Am contains a subword of Y-length 2 which is equal to a 0-word in G(0). But then the same subword occurs in A2,soA2 is not a shortest word representing g2, a contradiction. 2 3 (3) Notice that for every cyclic shift A1 of A, A1 is a subword of A .Itremains to apply part (2) of this lemma.

Lemma 3.5. — Suppose that R satisﬁes (Z1.1). Let A be a Y-shortest word representing an essential element g in G(0). Then the following conditions are equivalent: (i) A is cyclically Y-reduced; (ii) g is cyclically minimal in G(0) (that is g is not conjugate in G(0) of an element of a smaller Y-length); 2 (iii) g is cyclically Y-reduced in G(0) (that is |g |Y = 2|g|Y).

Proof. — (i)→(ii). Suppose that every cyclic shift of A is Y-reduced. Then A cannot be represented in the form UVU where |U|Y >0and U U is equal to a 0-word NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 69 in G(0). Therefore by Lemma 3.3 (2), A is a shortest word among all words representing conjugates of g.Henceg is not a conjugate of a shorter word, so g is cyclically minimal. 2 2 (ii)→(iii). Suppose that g is cyclically minimal but |g |Y <2|g|Y.ThenA is not a shortest word representing g2. By Lemma 3.3 (1), A2 is not Y-reduced, which means that A2 contains a subword W of Y-length 2 which is equal to a 0-word in G(0).Since A is reduced (by assumption it is a shortest word representing g), W ≡ U U where U is a sufﬁx of A, U is a preﬁx of A. By Lemma 3.1, |A|Y ≥ 2.HenceA ≡ UVU for some word V. The cyclic shift VU U of A is not Y-reduced, so it represents an element of G(0) which is shorter than g.Henceg is a conjugate of a shorter element, a contradiction. (iii)→ (i). This follows directly from Lemma 3.4 (3).

Lemma 3.6. — Suppose that R satisﬁes (Z1.1). Let x and g−1xg be 0-elements and g = hf where |g|=|f |+|h|. Then h−1xh is also an 0-element.

Proof. — Let X, G, H be Y-reduced words representing the elements x, g, h, respectively. By Lemma 3.3, the only possible reduction of the word G−1XG can be ful- ﬁlled in the middle of it. Then, after |h| such reductions, we will see that the subword HXH−1 is equal to a 0-word.

The next several lemmas deal with the subgroups 0(g) in property (Z3).

Lemma 3.7. — Suppose that R satisﬁes (Z1.1). Then: (1) the subgroup 0(g) is well-deﬁned, i.e. v0(u)v−1 does not depend on the presentation vuv−1 of an element g where u is cyclically y-reduced. (2) For every h ∈ G(0) and every essential g, we have 0(hgh−1) = h0(g)h−1.

Proof. — Assume that g = v u (v )−1 and v 0(u )(v )−1 = v0(u)v−1. Then

0(u ) = (v )−1v0(u)v−1v .

Thus, to prove the statement, one may assume that g is cyclically Y-reduced. There- fore one has to obtain equality 0(g) = v0(u)v−1 for the maximal 0-subgroups 0(g) and 0(u) normalized by g and u, respectively. By the symmetry, it suffices to prove that v0(u)v−1 is a 0-subgroup, because it is normalized by g = vuv−1. Let G, U, V be shortest words representing elements g, u, v. Then there is an annular diagram diagram over G(0) whose boundary labels are G ≡ φ(q1) and U ≡ φ( −1) . q2 and a path t, connecting the initial vertices of q1 and q2 is labeled by V Arbi- trary Y-band starting with q1 ends on q2, and vise versa, since g and u are cyclically Y-reduced. Since these words are essential, (q1)− can be connected with a vertex of q2 70 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR by a path p with label P where |P|=0. It defines a decomposition U ≡ U1U2. Hence = −1 k ( ) V PU1 U in G 0 for some integer k by Lemma 11.4 from [30]. k ( ) −k = ( ), ( ) = ( ). −1 ( ) Now, U 0 U U 0 U by the definition of 0 U 0 u Then U1 0 U U1 is also a 0-subgroup by Lemma 3.6 since the word U ≡ U1U2 normalizes 0(U). Fi- ( ) −1 = ( −1 ( ) ) −1 nally, V0 U V P U1 0 U U1 P is a 0-subgroup because P is a 0-word. The second statement of the lemma immediately follows from the first one.

Lemma 3.8. — Suppose that R satisﬁes (Z1.1). If g is an essential cyclically Y-reduced element of G(0) and x ∈ 0(g) then gsx is an essential and cyclically Y-reduced element for s = 0.

Proof. — Suppose that gsx = h−1uh for 0-element u and h ∈ G(0). Then the s m Y-length of (g x) is at most 2|h|Y for every integer m ≥ 0. On the other, we have s m sm (g x) = g x1 for some 0-element x1 ∈ 0(g) since the 0-subgroup 0(g) is normalized sm = −1 m −1 sm | | by g. Therefore g h u hx1 and the Y-length of g is at most 2 h Y for every integer m. This contradicts the fact that gsm is cyclically Y-reduced by the second part of Lemma 3.4. s s 2 s s Similarly if g x is not cyclically Y-reduced then |(g x) |Y <2|g x|Y = 2|g |Y. 2s 2s s 2 s Therefore for some 0-element x1 we have |g |Y =|g x1|Y =|(g x) | <2|g |Y which contradicts Lemma 3.4 (2).

Lemma 3.9. — Suppose that R satisﬁes (Z1.1). If g is an essential element and x ∈ 0(g) then 0(gx) = 0(g).

Proof. — By Lemma 3.7 we can assume that g is cyclically Y-reduced. Since x ∈ 0(g), the subgroup 0(g) which is normalized by g, is also normalized by gx.Since by Lemma 3.8, gx is also cyclically Y-reduced, we have 0(g) ⊆ 0(gx).Inparticular x ∈ 0(gx),sox−1 ∈ 0(gx). Therefore, as above, 0(gx) ⊆ 0(gxx−1) = 0(g).Hence 0(g) = 0(gx).

Lemma 3.10. — Suppose that R satisﬁes (Z1.1). If g is an essential element then for every integer m = 0, element gm is essential and 0(gm) = 0(g)

Proof. — By Lemma 3.7 we can assume that g is cyclically Y-reduced. By Lem- ma 3.4 (2) one may assume that gm is also cyclically Y-reduced. Then gm is essential by Lemmas 3.4 (2) and 3.3, and subgroups 0(g) and 0(gm) are 0-subgroups. It is clear that 0(g) ⊆ 0(gm).Letx ∈ 0(gm).Thenx ∈ 0(gms) for any s = 0. Take any integer t = 0.

We need to show that g−txgt is a 0-word. But we may choose s so that gms = gtgt for some t and |gms|=|gt|+|gt | by Lemma 3.3. Then g−txgt is a 0-word by Lemma 3.6.

Lemma 3.11. — Suppose that R satisﬁes (Z1.1). If g is an essential element of G(0) and x ∈ 0(g), l = 0,thenglx is an essential element and 0(gl x) = 0(g). NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 71

Proof. — The claim follows immediately from Lemmas 3.10, 3.9, 3.8 and 3.7. Lemma 3.12. — Suppose that R satisfies conditions (Z1.1), (Z1.2) and (Z3.2). Then for every essential element g ∈ G(0) which is cyclically minimal in rank 1/2,andeveryx ∈ 0(g), we have (gx)n = gn. Proof. — By the definition of 0(g) and Lemma 3.7, we can assume that g is cyclically reduced and x is a 0-element. We have (gx)n = gn(g−n+1xgn−1)...(g−1xg)x.Byby property (Z3.2), there exists a 0-element r such that gr−1 commutes with every element of 0(g). Therefore r also normalizes 0(g) and the actions of r and g on 0(g) (by conjugation) coincide. Hence for every integer s the actions of rs and gs on 0(g) also coincide, so g−sxgs = r−sxrs. Therefore (g−n+1xgn−1)...(g−1xg)x = (r−n+1xrn−1)...(r−1xr)x = r−n(rx)n. By property (Z1.2) (rx)n = rn = 1 in G(0).Hence(gx)n = gn. Lemma 3.13. — Suppose that R satisfies conditions (Z1.1), (Z1.2), (Z3.2). Then for every essential element g ∈ G(0) which is cyclically minimal in rank 1/2,andeveryx ∈ 0(g) we have gnx = xgn. Proof. — It follows from Lemma 3.12 that gn commutes with both g and gx. Therefore it commutes with x. Lemma 3.14. — Suppose that R satisfies (Z1.1). Let A be a Y-reduced essential word over X (i.e. it represents an essential element in G(0))andletC be a 0-word. Suppose that the subgroup generated by A−sCAs for all integers s consists of 0-elements. Then C ∈ 0(A). Proof. — If A is cyclically Y-reduced then C ∈ 0(A) by definition. Otherwise by Lemma 3.3 (2), A can be represented as UVU where U U is equal to a 0-word W in G(0) and V is cyclically Y-reduced, so every cyclic shift of V is Y-reduced. Since we assume that A is essential, VW is an essential word. By Lemma 3.3 (1) then every cyclic shift of VW is Y-reduced. Hence VW is cyclically Y-reduced. By Lemma 3.6, U−1CU is a 0-element. Since the subgroup generated by A−sCAs, s ∈ Z consists of 0-elements and is normalized by A, U(VW)−s(U−1CU)(VW)s)U−1 is a 0-element for every s. Notice that U(VW)sU−1 is a Y-reduced form of As. Indeed, V is cyclically Y- reduced subword of A by definition, VW is a conjugate of A of the same Y-length as V. Hence by Lemma 3.3 (2), VW is cyclically Y-reduced. By Lemma 3.4 (2) then (VW)s is cyclically Y-reduced. By Lemma 3.3(1), U(VW)sU−1 is reduced because every subword of Y-length 2 of U(VW)sU−1 is either a subword of A or a subword of (VW)s. Since A−sCAs = U(VW)−s(U−1CU)(VW)s)U−1 is a 0-element for every s, and U ≡ WU−1 is a suffix of As, by Lemma 3.6, the subgroup generated by 72 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

(VW)−s(U−1CU)(VW)s consists of 0-elements and is normalized by U−1AU = VW. Hence U−1CU is in 0(U−1AU). Therefore C ∈ 0(A) by Lemma 3.7.

Lemma 3.15. — Suppose that R satisﬁes (Z1.1), (Z1.2), (Z3.1) and (Z3.2). Let g be an essential element of G(0) represented by a cyclically Y-reduced word A in G(0). Suppose that A contains a subword B6,whereB is a cyclically minimal in rank 1/2 word of a positive Y-length. Let A be the word obtained by replacing that occurrence of B6 by B6−n.Then0(A) ≤ 0(A ), provided the word A represents an essential element of G(0).

Proof. — Let x ∈ 0(A). Then for every integer s, g−sxgs is a 0-element. Fix s>0. Let C be a 0-word representing x, D be the word representing g−sxgs.NoticethatB6 contains the ﬁfth power of every cyclic shift of B.LetB1 be a cyclic shift of B starting ≡ 5 with a letter from Y.LetA A1B1A2. Clearly then the word A obtained from A by 6 6−n 5−n replacing B by B is equal to A1B1 A2 inthefreegroup. ( ) −1 −1 Consider a van Kampen diagram over G 0 with boundary label p1q1p2 q2 s where φ(p1) = C, φ(p2) = D, φ(q1) = φ(q2) = A .Thepathq1 can be represented as

q1 = t1,1t1,2t1,3t1,4t2,1t2,2t2,3t2,4...ts,1ts,2ts,3ts,4

φ( ) ≡ 4 φ( ) ≡ = , ..., φ( ) 5 where ti,2 B1, ti,3 B1, i 1 s ( ti,2ti,3 is the ﬁxed occurrence of B1 in A). s Since A is Y-reduced and p1, p2 do not contain Y-edges, the maximal Y-bands starting on q1 provide a one-to-one correspondence between Y-edges of q1 and Y- edges of q2.SinceB1 starts with a Y-letter, ti,2 and ti,3 start with a Y-edges, ei and fi respectively, i = 1, ..., s. For every i = 1, ..., s, consider the maximal Y-band Ti starting on ei and the maximal Y-band Bi starting on fi. Consider the smallest subdiagram ∆i bounded by asideofTi,asideofBi, q1, q2, containing Ti but not containing Bi. The boundary ∆ ( )−1( )−1 φ( ) ≡ φ( ) ≡ 4 ≡ φ( ) of each i has the form piti,2 pi ti where ti,2 ti B1, vi pi and ≡ φ( ) vi pi are 0-words. Therefore −4 4 = B1 viB1 vi for every i = 1, ..., s. By Property (Z3.1) (all premises of (Z3.1) hold), vi ∈ 0(B1). By Lemma 3.13, vi n = ,..., Γ commutes with B1.Foreveryi 1 s consider a van Kampen diagram i over ( ) ¯ ¯ ( ¯ )−1(¯ )−1 φ( ¯ ) ≡ φ( ¯ ) ≡ φ(¯ ) ≡ φ(¯ ) ≡ −n G 0 with boundary piqi pi qi where pi pi vi, qi qi B1 . ∆ Now cut the diagram along each path pi.Thuseachpathpi turns into two copies of pi, which we shall call the left copy and the right copy: the right copy is contained ¯ in the diagram containing the band Ti.Foreveryi = 1, ..., s, identify paths pi of Γ ¯ = ,..., i with the left copy of pi, pi with the right copy of pi, i 1 s.Asaresult,we ( −n 5 )−s ( −n 5 )s −1 get a van Kampen diagram with boundary label A1B1 B1A2 C A1B1 B1A2 D . NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 73

5−n = ( )−s ( )s = ( ) Sinceaswehavenoticed,A1B1 A2 A ,wehavethat A C A D in G 0 for every positive s. The same result for negative s can be proved similarly. Hence the subgroup generated by (A )−sC(A )s, s ∈ Z, consists of 0-elements and is normalized by A . By Lemma 3.14, C ∈ 0(A ) as required.

Lemma 3.16. — Suppose that R satisﬁes (Z1.1), (Z2.1), (Z2.2), (Z2.3). Let g be an essential element of G(0) represented by a cyclically Y-reduced word A in G(0). Suppose that A contains a subword B which is a preﬁx of a cyclic shift C of a hub, C ≡ BB ,and|B|≥ε|C|.Let A be the word obtained by replacing that occurrence of B by (B )−1. Then, provided A represents an essential element of G(0), we have inclusion 0(A) ≤ 0(A ).

Proof. — The proof is similar to the proof of Lemma 3.15. Let x ∈ 0(A).Then for every integer s, g−sxgs is a 0-element. Fix s>0.LetC be a 0-word representing x, D be the word representing g−sxgs. Since letters not from Y have Y-length 0,wecan assume that B starts and ends with Y-letters. Let A ≡ A1BA2. Clearly then the word −1 −1 A obtained from A by replacing B by (B ) is equal to A1(B ) A2 in the free group. −1 −1 Consider a van Kampen diagram with boundary label p1q1p2 q2 where s φ(p1) ≡ C, φ(p2) ≡ D, φ(q1) ≡ φ(q2) ≡ A .Thepathq1 can be represented as

q1 = t1,1t1,2t1,3t2,1t2,2t2,3...ts,1ts,2ts,3 where φ(ti,2) ≡ B, i = 1, ..., s (φ(ti,2) is the ﬁxed occurrence of B in A). s Since A is Y-reduced by Lemma 3.4 (2), and p1, p2 do not contain Y-edges, the maximal Y-bands starting on q1 provide a one-to-one correspondence between Y- edges of q1 and Y-edges of q2.SinceB starts and ends with a Y-letter, ti,2 starts and ends with a Y-edges, ei and fi respectively, i = 1, ..., s. For every i = 1, ..., s, consider the maximal Y-band Ti starting on ei and the maximal Y-band Bi starting on fi. Consider the smallest subdiagram ∆i bounded by asideofTi,asideofBi, q1,andq2, containing Ti and Bi. The boundary of each ∆ ( )−1( )−1 φ( ) ≡ φ( ) ≡ ≡ φ( ) ≡ φ( ) i has the form piti,2 pi ti where ti,2 ti B, vi pi and vi pi are 0-words. Therefore = viB Bvi for every i = 1, ..., s. By Property (Z2.3) (all premises of (Z2.3) hold), we have ( )−1 = ( )−1 , = , ..., . vi B B vi i 1 s

Thus for every i = 1, ..., s there exists a van Kampen diagram Γi over R with ¯ ( )−1(¯ )−1 φ(¯ ) ≡ ( )−1 φ(¯ ) ≡ ( )−1 boundary piti,2 pi ti where ti,2 B , ti B . Now we can substitute the subdiagrams ∆i in ∆ by Γi. The boundary label of the resulting diagram is equal (A )−sC(A )sD−1.Hence(A )−sC(A )s = D in G(0) for every positive s.The same result for negative s can be proved similarly. 74 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Hence the subgroup generated by (A )−sC(A )s, s ∈ Z, consists of 0-elements and is normalized by A . By Lemma 3.14 C ∈ 0(A ) as required.

3.3. The construction of a graded presentation

In this section, we show how the theory from [30, Chapter 6] can be adapted to diagrams over a presentation satisfying (Z1), (Z2), (Z3). We are defining several concepts analogous to concepts in [30], Chapter 6, by induction on the rank i. We say that a certain statement holds in rank i if it holds in the group G(i) given R =∪i S by i j=0 j. Recall that a word w is A-periodic if it is a subword of some power of the word A. Any A-periodic word can be read (say, clockwise) on a cyclic graph with |A| edges, labeled, starting with a vertex o, by the word A. Then every A-periodic word corresponds to a path on this graph. Let W ≡ W1W2 be a decomposition of an A-periodic word W, such that the word W1 ends at o, when reading the word W on the cyclic graph, |W|≥|A|. Then we say that the decomposition is a phase decomposition of W. Ifthelabelofasimplepathp of an arbitrary labeled graph has an A-periodic label W ≡ φ(p), then a vertex of p, defining a phase decomposition of p is said to be a phase vertex of p. Up to the end of this section, the Y-length of a word W will be called simply length and will be denoted by |W|. Letusdefinesimplewordsofranki = 0, 1/2, 1, 2, ... and periods of rank i = 1, 2, 3,....LetS0 = R0 and S1/2 be the set of hubs. Thus the groups G(0) and G(1/2) = G are defined by relations of the sets R0 and R1/2 = S0 ∪ S1/2, respectively. A word A of positive length is said to be simple in rank 0 (resp. 1/2) if it is not a conjugate in rank 0 (resp. 1/2) either of a word of length 0 or of a product of the form BlP,where|B| < |A| and P ∈ 0(B) in rank 0.

Suppose that i ≥ 1, and we have defined the sets of relators Rj, j1, 1/2 if i = 1 or 0 if i = 1/2. Similarly i+ is 1/2 if i = 0 or 1 if i = 1/2 or i + 1 if i ≥ 1. ≥ X For every i 1 consider the set i of all words of length i which are simple ∼ in rank i−, and the equivalence i− given in Lemma 3.24 (we apply it for the smaller X X rank here). Now choose a set of representatives i of the equivalence classes of i . The words of Xi are said to be periods of rank i. The set of words Ri (defining the ( ) R S ={ n, ∈ X } group G i or rank i)istheunionof i− and i A A i . Let A be a word of positive length. We say that A is simple in rank i ≥ 1 if it is not a conjugate in rank i either of a word of length 0 or of a word of the form BlP, NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 75 where B is a period of rank j ≤ i or an essential word with |B| < |A|,andP represents a word of the subgroup 0(B). R(∞) =∪∞R Π R(∞) Let 0 i.If is a cell of a diagram over with the boundary labeled by a word of the set Sj, then, by definition, r(Π) = j. The boundary label of every cell of a rank ≥ 1 (read counterclockwise), has a period A defined up to cyclic shifts.

For j ≥ 1, a pair of distinct cells Π1 and Π2 of rank j of a diagram ∆ is said to n −n be a j-pair, if their counterclockwise contours p1 and p2 are labeled by A and A for aperiodA of rank j and there is a path t = (p1)− − (p2)− without self-intersections n such that φ(t) is equal, in rank j−,toanelementof0(A). Then φ(t) and A commute −1 in rank j− by Lemma 3.13, and so the subdiagram with contour p1tp2t can be replaced in ∆ by a diagram of rank j−.Asaresult,weobtainadiagramwiththesame boundary label as ∆ but of a smaller type. Similarly, a pair of hubs Π1, Π2 with counterclockwise contours p1, p2 forms 1 ( )− ( )− a 2 -pair,ifthevertices p1 and p2 are connected by a path t without self-inter- −1 sections such that the label of the path p1tp2t is equal to 1 in G(0). Consequently, any diagram can be replaced by a diagram having no j-pairs, j = 1/2, 1, 2,..., i.e. by a g-reduced diagram with the same boundary label(s)8. Assume that q1 and q2 are disjoint sections of the boundary ∂(∆) of a dia- −1 gram ∆ of rank i, and φ(q1), φ(q2) are A-andA -periodic words respectively, where |q1|, |q2|≥|A|, A is simple in rank i or a period of rank j ≤ i.Wecallq1 and q2 compatible if there exist phase vertices o1 and o2 on q1 and q2, respectively, and a path t = o1 − o2 without self-intersections, such that φ(t) is equal in rank i to an element of 0(A); moreover the equality must be true in rank j− if A is a period of rank j. Similarly one deﬁnes the compatibility of an A-periodic cell Π of rank j ≥ 1 with a A±1-periodic section q of ∂(∆).

Lemma 3.17. —LetΠ1 be a cell of rank j ≥ 1 in a g-reduced diagram ∆ of rank i. n Assume it has a boundary label A .LetΓ be a subdiagram of ∆ with contour p1q1p2q2,where q1 is a subpath of ∂(Π1) and q2 is a subpath of the contour of a cell Π2 with the boundary label −n A ,(Π1, Π2 do not belong to Γ). Then

1) q1 and q2 cannot be compatible in Γ; 2) the subdiagram Γ contains no cells compatible with q1.

Proof. — If one of the statements were false, we could include the cell Π1 into a j-pair and come to an obvious contradiction with the assumption that ∆ is g-reduced diagram, as in Lemma 13.2 from [30].

8 Recall that in [30], §13. 2, such diagrams are called reduced. But in this paper we shall consider two kinds of reduced diagrams, so we call diagrams without j-pairs g-reduced; “g” stands for “graded”. 76 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Lemma 3.18. —Letp1, p2... be a partition of the boundary of a diagram ∆ of rank l i into several sections, and φ(p1) ≡ A for some period of rank j. Assume that a cell Π of ∆ is compatible with p1. Then there exists a g-reduced diagram ∆ with topologically identical partition of , , ..., τ(∆ ) τ(∆), φ( ) ≡ φ( ) ≥ ,φ( ) ≡ l+sn the boundary into p1 p2 such that < pk pk for k 2 p1 A for some integer s, and ∆ has no cells compatible with p1.

Proof. — Let t be the path connecting a phase vertex o1 of p1 and a phase vertex ±1 o2 of the A -periodic cell Π, according to the definition of the compatibility, and T ≡ φ(t) is equal in rank i to an element of 0(A). The vertex o1 defines a decomposition l1 l2 l1 ±n −1 l2 A A , and there is a path p homotopic to p1 with label W ≡ A TA T A . By Lem- l±n ma 3.13, we have W = A in rank j−. Thus one can cut Π out of ∆, by cutting along p. Then one can replace the deleted subdiagram by the diagram of the equality W = Al±n. By continuing this process, we prove the lemma. The following proposition is the main statement of this section. Proposition 3.19. — If the presentation R of a group G satisfies properties (Z1), (Z2), (Z3) then the graded presentation S0 ∪ S1/2 ∪ S1 ∪ S2 ∪ ... satisfies statements of Lemmas 3.20 through 3.39 below and is such that R = S0 ∪ S1/2. In particular, all g-reduced diagrams over n this presentation are A-maps. Every relator from Si, i ≥ 1,isoftheformA for a cyclically reduced word A of Y-length i. The group defined by this presentation is G/gn, g ∈ G. Lemmas from 3.20 to 3.39 will be proved by simultaneous induction ≥ 1 . on rank i 2 Lemma 3.20. —LetV be a periodic word with a simple in rank i period. Assume that for some word U with 2|U|≤|V|,thewordVU is a conjugate in rank i of a 0-word. Then there is a decomposition V ≡ V1V2, such that ||V1|−|V2|| < β|V|+|U| and for some word S of length at most 3γ |V|, V2UV1 = S in rank i. Proof. — Consider a g-reduced annular diagram ∆ for the conjugacy of VU to a0-wordW. We may assume that ∆ has the smallest possible type. Therefore there are no Y-bands surrounding the hole of ∆, since otherwise their boundary labels were 0-words. Hence, by Lemmas 3.38 and 2.16, there are vertices o1 on the contour q1 labeled by VU and o2 on the contour q2 labeled by W, which can be connected by asimplepatht of length at most γ(|V|+|U|) ≤ 3γ |V|/2 (see Figure 7). One may suppose that the vertex o2 does not cut the word W, because |W|=0. The vertex o1 cannot cut the word |U|. Indeed, in case it defines a decomposition −1 −1 −1 U ≡ U1U2, we obtain the following equality in rank i: U1TW T U2 = V , where T ≡ φ(t). It follows by Lemmas 3.39 and 2.7 that β¯|V| < |U|+2γ(|V|+|U|), whence |V| < (β¯ − 2γ)−1(1 + 2γ)|U|≤2|U|, contrary to the hypothesis of the lemma. −1 −1 Thus the vertex o1 defines a decomposition V ≡ V1V2, and TW T V2U = −1 . β¯ | | | |+| |+ γ(| |+| |) | |− V1 in rank i Then, as above, V1 < V2 U 2 V U . Therefore V1 NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 77

V1 o1 t o2 U W V2

FIG. 7.

|V2| < β|V1|+|U|+3γ |V|. Similarly, |V2|−|V1| < β|V1|+|U|+3γ |V|, and so ||V1|−|V2|| < β|V|+|U|,as2|U|≤|V|. To complete the proof we observe that the product V2UV1 is equal in rank i to the word S ≡ TWT−1 of length at most 3γ |V|.

±1 Let q1, q2 be two paths labeled by A -periodic words for some period A. We say that vertices o1, o2 ofthesepathsareinthesame phase (with respect to A)ifthey are phase vertices for the paths q1, q2 with respect to a cyclic permutation A of the word A.

Lemma 3.21. —LetΠ be a cell of positive rank in a g-reduced disc diagram ∆ of rank −1 i with contour p1q1p2q2,whereq1 and q2 are labeled by A-periodic words where A is a simple in rank i word. Suppose there are disjoint contiguity subdiagrams Γj of Π to qj of degrees ψj for = , , ψ + ψ − α/ . j j j j = ∂(Π, Γ , ) j 1 2 with 1 2 >1 2 Denote s1 t1 s2 t2 j qj (see Figure 8). Let p be −1, = ( 2) , = − a subpath of q2 such that vertex o s2 + belongs to p where the lengths of subpaths p p− o 1 = − + |∂(Π)|. and p o p are less than 3 Then any vertex o2 of p is not in the same phase with = ( 1) vertex o1 s1 −.

Proof. — By Lemmas 3.38, 3.39 and 2.3,

(10) ψ1,ψ2 ∈[1/2 − 2α, α¯], and by Lemma 2.2, | 1|, | 2|∈[( + β)−1( / − α)|∂(Π)|,( − β)−1α¯|∂(Π)|]. (11) t2 t2 1 2 1 2 2 1 2

Assume that o2 and o1 are in the same phase and o2 ∈ p . First, let Π be a hub. (Γ ) = (Γ ) = | j|= , = Then r 1 r 2 0 by Lemma 3.37 in the smaller rank, hence sl 0, l j 1, 2. Recall that by (Z2.2), every hub is a linear word. By deﬁnition, every simple word in rank i is cyclically Y-reduced. j Therefore by Lemmas 3.2 and 3.4 every maximal Y-band starting on tl ends on j j t3−l and forms a 0-bond between these paths. Hence any subpath of tl starting and j ending with Y-letters, is a contiguity arc for some contiguity subdiagram between tl 78 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

j , = , . 1 and t3−l for l j 1 2 By our assumption and inequality (11), there are subpaths u 2 1 ( 2)−1 and u of t2 and t2 with the same label U,suchthat 1 |U| > ((1 + 2β)−1(1/2 − 2α) − 1/3)|∂(Π)| > |∂(Π)|. 7 Gluing along u1 and u2 one can construct a contiguity diagram of rank 0 between Π 1 |∂(Π)|. disjoint arcs of the cell of length greater than 7 But this contradicts condition (Z2.2).

t1 o q1 ... 2 1 ... 1 x1 y2 2 1 1 Γ1 x1 s1 s2 Γ y1 1 1 1 1 t1

t2 1 y2 2 2 1 Γ x2 2 s1 x1 y2 2 2 Γ2 s2 ... 2 ... 2 q2 t2 p− o2 o p+

FIG. 8.

Now assume that Π is a cell with a period B of rank k ≤ i.Inthiscase, r(Γ1), r(Γ2) 7 Lemma 2.1 and inequality (10) show that Lemma 2.12 is applicable to each Γ , Γ . | j| ζ|∂(Π)| of the diagrams 1 2 Since s1 < , it follows from Lemma 2.12 that an j , ∈{, } Γ j arbitrary vertex of tl (l j 1 2 ) can be connected in j with a vertex of t3−l by apathoflengthatmost3γ −1ζ|∂(Π)|. Therefore, for j = 1, 2, there exist subdiagrams Γ Γ j j j j, | j| γ −1ζ|∂(Π)|,φ( 1) ≡ φ( 2)−1 ≡ j of j with contours x1 y1x2 y2 where xl <3 y2 y2 U and j j. y1 are subpaths of t1 Γ Γ 1 ( 2)−1, By gluing diagrams 1 and 2 along y2 and y2 one can obtain (after delet- Γ 1 2 | | | | ing j-pairs) a g-reduced diagram 0 with contour z1 y1z1 y1 where z1 , z2 < 6γ −1ζ|∂(Π)|. Notice, that by property A1 and Lemma 2.7, 1 | y j|≥β¯ | y j|−|x j|−|x j| > β¯|∂(Π)|−6γ −1ζ|∂(Π)| 1 2 1 2 7 1 n > |∂(Π)|≥ |B|. 8 8 NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 79 Γ Γ Γ Therefore one can apply Lemma 2.15 to 0 and obtain a subdiagram 0 of 0 with , j = , | |, | | α| |, contour v1w1v2w2 where wj is a subpath of y1 (for j 1 2), v1 v2 < B and

1 n |w |, |w | > |∂(Π)|−γ −1(12γ −1ζ|∂(Π)|+|B|) > |B| 1 2 8 9 since |∂(Π)|≥n|B| and the rank of Π is k =|B|.Sincethelabelsofbothw1 and w2 (Γ ) , are B-periodic words, and period B is simple in rank r 0

Lemma 3.22. —Let∆ be a narrow contiguity diagram of rank i between q1 and q2 with , −1 contour p1q1p2q2 where q1 and q2 are paths with A-periodic labels for some simple in rank i period A. Assume that there are vertices o1 and o2 on q1 and q2 respectively, which are in the same phase, the label φ(o1 − o2) of a path o1 − o2 is a conjugate of a 0-word in rank i and 1 ||( )− − |−|( )+ − || ≤ (| |, | |), ( )− − − ( )+ q1 o1 q2 o2 3 min q1 q2 where q1 o1 and o2 q2 are subpaths of q1 and q2.Thenr(∆) = 0.

Proof. — As in the proof of Lemma 2.13, ∆ can be partitioned into several narrow bonds El of an α-series and several subdiagrams ∆1, ∆2, ... between successive bonds. If r(El ) = r(El+1) = 0, then r(∆l ) = 0 by Lemma 2.11 and the maximality of the α-series. Thus it sufﬁces to prove that r(El ) = 0 for every El. Proving by contradiction, we choose the bond El whose principal cell Π0 has maximal perimeter (see Figure 9).

q1 o1 q1 o 1 q t q1 1 s1 2 s1 2 1 1 t1 t p2 p1 u1 u2 Π0 ∆ 2 t1 2 2 s s2 o 1 2 q2 t2 q2 o2 v1 v2

FIG. 9. 80 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

¯ −1 By Lemma 2.7 (and Lemmas 3.38, 3.39) |q2|≤β (|q1|+|p1|+|p2|). We have |p1|+|p2| < (α + 4ζ)|∂(Π0)| as in Lemma 2.13 for the side arcs of narrow bonds. Denote by Γ1 and Γ2 the contiguity subdiagrams of Π0 to q1 and q2 with contours j j j j = , Γ | | s1 t1 s2 t2 (j 1 2) as in Lemma 3.21. Applying Lemmas 2.2, 2.3, to 1,wehave q1 > ( − α − α)¯ |∂(Π )|, 1. | |+| | 5 α| | | | 1 0 because q1 contains t2 Hence p1 p2 < 2 q1 and q2 < (1 + 3α)|q1|. = 1 = 2 ∂(Π) = 1 2 Then we decompose q1 q1t2 q1, q2 q2t2 q2 and t1 u1t1 u2. Observe that since El is also a bond of a narrow contiguity subdiagram between 1 2 , | | | | the paths q1t2 and t2 q2 we have the similar inequality as for q1 and q2 above: | 1| ( + α)| 2 |. | |+| | ( + α)(| |+|2|+| |) = q1t2 < 1 3 t2 q2 This implies that q1 q2 < 1 3 q2 t2 q2 ( + α)| |. | |+| | ( + α)| |. 1 3 q2 Similarly, q1 q2 < 1 3 q1 ∈ ∈ Assume ﬁrst that o1 q1 and o2 q2. We may also suppose by symmetry that | | 1+3α (| |, | |). , q1 < 2 min q1 q2 Notice that equal shifts of the vertices o1 o2 along q1 and q2 preserve the assumption of the lemma. Now assume that, after shifts, o2 becomes ( ) equal to q2 − and o1 still belongs to q1. The last assumption leads to a contradiction, 1+3α | −( )+|−| −( )−| | |− | | becausewehaveforthenewo1 and o2 that o2 q2 o1 q1 > q2 2 q2 > | |/ ( 1) q2 3 contrary to the assumption of the lemma. Therefore one can move o1 to t2 − and o2 to a vertex laying on q2,inthiscase. ∈ ∈ 2 ∈ If o1 q1 and o2 t2 (o2 q2), then obviously, one can make equal shifts of these ( 1) ( 2) vertices so that either o1 becomes equal to t2 − or o2 becomes equal to t2 −. Similarly, , ( 1) ,( 2) one of the vertices o1 o2 can be made equal to one of the t2 ± t2 ± after parallel ∈ 1 ∈ 2. shifts, if o1 t2 and o2 t2 The remaining cases are symmetric to the examined ones. = ( 1) = ( 1) , Thus, to prove the lemma, one may assume that either o1 t2 − or o1 t2 + = ( 2) , = ( 2) . = ( 1) = ( 1) . or o2 t2 − or o2 t2 + We will assume that o1 t2 + s1 − 2 , In addition, one may assume that the vertex o2 belongs to the subpath t2 q2 be- 2 1 | − ( )−|≥ |∂(Π )| cause otherwise it belongs to q2 and o2 t2 3 0 by Lemma 3.21. In this ( 1) case, by Lemma 2.2, one might shift o1 to t2 − so that corresponding vertex o2 shifts 2 to a vertex of the path q2t2 because

2 1 2 | − ( )+| |∂(Π )|+| | ( / + ( − β)( − α − α))¯ |∂(Π )| o2 t2 > 3 0 t2 > 1 3 1 2 1 0 > 2 |∂(Π )| ( + β)α¯|∂(Π )| | 1|. 3 0 > 1 2 0 > t2 Then one gets a situation which is symmetric to the one considered above with respect → −1 → −1 → −1 → −1 ∆ to changing of notations p1 p2 , p2 p1 , q1 q1 , q2 q2 (i.e. we replace by its mirror copy). = ( 1) Thus, up to symmetries, one has to consider only one case when o1 t2 + and ∈ 2 . , = − ( 2) o2 t2 q2 The vertices o2 and o1 can be connected by a path vu where v o2 t2 − = ( 2)−1 ( 1)−1. ≡ φ( ) ≡ φ( ). | |≤ and u s2 u2 s1 Set V v and U u As has been mentioned, U (α + ζ)|∂(Π )|. | |≥ 1 |∂(Π )| 2 0 On the other hand, V 3 0 by Lemma 3.21. Therefore one gets the decomposition V ≡ V1V2 guaranteed by Lemma 3.20. Denote by o the vertex of v which gives the corresponding decomposition v = v1v2. There are two subcases. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 81

−1 (1) |V| <3α |∂(Π0)|. Recall, that by Lemma 3.20 the word V2UV1 is equal in rank i to a word S of length at most 3γ |V| and

−1 (12) ||V1|−|V2|| < β|V|+|U| < (3α β + α + 2ζ)|∂(Π0)| <2α|∂(Π0)|.

By Lemma 3.21, |V|≥|∂(Π0)|/3, whence |V1|, |V2| > (1/6−α)|∂(Π0)| >1/7|∂(Π0)| by (12). Since

. |∂(Π )| ( + β)−1( − α¯ − α)|∂(Π )| | j| 0 45 0 < 1 2 1 0 < t2 −1 < (1 − 2β) α¯|∂(Π0)| ∈ 2 by Lemmas 2.3 and 2.2, we have in case o t2 that | − ( 2) | | 2|−| | (( + β)α¯ − / + α)|∂(Π )| o s1 − < t2 V2 < 1 2 1 6 0

<0. 35|∂(Π0)|. ∈ If o q2,then | − ( 2) |=| |−|2| | |+ α|∂(Π )|− . |∂(Π )|. o s1 − V2 t2 < V1 2 0 0 45 0 ( 2) ( . |∂(Π )|, Hence in any case the path between o and s1 − has length at most max 0 35 0 | |− α|∂(Π )|), α . . | 2 −1 1| V1 2 0 because of (12) and the inequality 4 <045 Since s1u1 s2 < (α + ζ)|∂(Π )|, ( 1) 2 0 the vertex o can be connected with t2 − by a path p of length at most max(0. 4|∂(Π0)|, |V1|). 1 . The path pt2 is homotopic to the path v2u Thus the word S is equal in rank i φ( 1) ≡ φ( ). ≡ φ( 1) to the word P t2 V1,whereP p The word W t2 V1 is A-periodic because | |+| 1| | |+ the vertices o2 and o1 are in the same phase. It has length at least V1 t2 > V1 −1 0. 45|∂(Π0)|. Then W = P S in rank i, which implies, by Lemmas 2.7 and 3.38, 3.39 applied to the diagram for this equality, that

( . |∂(Π )|, | |) + γ | | | |+| | β(¯ | |+|1|) max 0 4 0 V1 3 V > P S > V1 t2 (13) ¯ > β(|V1|+0. 45|∂(Π0)|).

−1 On the other hand, since |V1|≤|V|≤3α |∂(Π0)|, we have (β + 3γ)|V| < ¯ (0. 45β − 0. 4)|∂(Π0)| which contradicts inequality (13). −1 (2) Let |V|≥3α |∂(Π0)|. Recall that |V1|−|V2| < β|V|+(α + 2ζ)|∂(Π0)|, | |=| |+| |≥ α−1|∂(Π )| 7 | |≥ and since V V1 V2 3 0 , the length of v2 is greater than 15 V . α−1|∂(Π )| ∈ 1 1 4 0 . By Lemma 2.13, the vertex o can be connected with a vertex o q1t2 −1 1 by a path t of length at most α |∂(Π0)| < |V|. By Lemma 2.7, the length of the 3 path q = o − o1 is at least 7 1 β¯|v |−|t|−|u| > |V|− |V|−α2|V| > |V|/8 2 15 3 82 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

2 because |u|≤(α + 2ζ)|∂(Π0)| < α |V| by the assumption of case (2). Therefore the length of the A-periodic word W ≡ QV1 (where Q ≡ φ(q))isgreaterthan ( 7 + 1 )| |≥ 7 | |. 15 8 V 12 V By Lemma 2.7, W cannot be equal in rank i to a word of length ≤|V|/2. But it is equal in rank i to the label of t times S which is equal in ( 1 + γ)| |, rank i to V2UV1. Therefore W is equal in rank i to a word of length < 3 3 V a contradiction.

Lemma 3.23. — Assume that A is a word which is not a conjugate in rank i of a 0-word. Let A = XBlRX−1 in rank i,whereB is a simple in rank i word or a period of rank j ≤ i and R ∈ 0(B).Then0(A) ≤ X0(B)X−1 in rank i (i.e. the canonical image of 0(A) in G(i) is asubgroupoftheimageofX0(B)X−1).

Proof. — One may suppose that A is a Y-reduced word, moreover that A is cyclically Y-reduced, by Lemma 3.4(1) and Lemma 3.7. l −1 Consider a diagram ∆1 of rank i for the equality A = XB RX .Let∆2 be an annular diagram which arises after identifying the boundary sections of ∆1 labeled l by X.Ithascontoursq1 and q2 labeled by A and B R, respectively. Initial vertices o1 of q1 and o2 of q2 are connected in ∆2 by a path x labeled by the word X. If ∆2 is not g-reduced, one can reduce it and obtain a diagram ∆3 of a smaller type in which the vertices o1 and o2 are connected by a path x with X ≡ φ(x ) = X in rank i. (Lemma 11.3 [30].) We will induct on the type of ∆3. l −1 If r(∆3) = 0, then 0(A) = X 0(B R)(X ) in rank 0, by Lemma 3.7. Since X = X in rank i, the statement of the lemma follows from Lemma 3.11. Assume now that r(∆3) >0. One may suppose that there are no cells compatible with q2 in ∆3 (if B is a period of rank j ≤ i), since otherwise one can decrease ±sn thetypeof∆3 by Lemma 3.18 and replace the label of q2 for B R. Therefore the cyclic section q2 is smooth by Lemma 3.39, and so the degree of contiguity of any cell of ∆3 to q2 is less α¯ by Lemma 2.3. By Lemma 2.4 there is a positive cell Π and its Γ (Π, Γ , ) 1 (γ¯ − α)¯ ε. contiguity subdiagram to q1 such that q1 > 2 > It follows, from Lemma 2.6 that there is a positive cell π of ∆3 and a contiguity subdiagram Γ of rank 0 of π to q1, such that (π, Γ, q1) ≥ ε. Let ∂(π, Γ, q1) = s1t1s2t2. Then |s1|=|s2|=0 since r(Γ) = 0.Thereare no Y-bands connecting different edges of subpath t1 of ∂(π) by axiom A2. Therefore

|t1|≤|t2|, and the word A obtained by the exchange in A of the subword φ(t2) by −1 φ(s2t1s1) , is Y-reduced (and equal to A in rank 0.) −1 −1 The subword φ(t1) of A is a subword of the boundary label of π and |φ(t1) | −1 ≥ ε|∂(π)|. By Lemmas 3.15, 3.16, if one substitutes φ(t1) in A by φ(t),wheret1t is the contour of π, and replaces the resulting word by an equal in rank 0Y-reduced word A , then 0(A ) ≥ 0(A ) = 0(A).

The word A is the label of the contour of a g-reduced diagram ∆4 which is obtained from ∆3 by cutting the cell π out and (possibly) adding several 0-cells. Of NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 83 course, such surgeries (cutting off a contiguity diagram or a cell) can touch the path x , but there is a homotopic path x with label X = X = X in rank i which is not touched by the surgeries (i.e. this path does not pass through the contiguity subdiagram and does not have common edges with the cell). By the inductive hypothe- −1 sis for the diagram ∆4 of a smaller type, 0(A ) ≤ X 0(B)(X ) in rank i.Hence 0(A) ≤ X0(B)X−1, as desired.

We need the statement of the following lemma to deﬁne periods of rank i+.

Lemma 3.24. — The following relation ∼i is an equivalence on the set of all simple in rank i words: A ∼i B by deﬁnition, if there are words X, P, R,whereP ∈ 0(A), R ∈ 0(B) (in rank 0), such that AP = XB±1RX−1 in rank i.

Proof. — Obviously, ∼i is a reﬂexive relation. The equality AP = XB±1RX−1 implies B±1R = X−1APX in rank i.Taking −1 ±1 the inverses, if necessary, we have equality BR± = X A P±X for a suitable P± ∈ 0(A), R± ∈ 0(B),because0(A) and 0(B) are normalized by A and B respectively. −1 By Lemma 3.9 0(B) = 0(BR±),andsoR± ∈ X 0(A)X by Lemma 3.23. Then −1 −1 ±1 Q ≡ XR±X ∈ 0(A) by Lemma 3.7. Hence B = X A VX in rank i,where V ≡ P±Q ∈ 0(A), and so the relation ∼i is symmetric. −1 ±1 We have also proved that B ∼i A implies that B = X A VX in rank i for some V ∈ 0(A) and a word X. ±1 −1 Assume now that B ∼i C,thatisB = YC QY in rank i for a word Y and ±1 −1 −1 some Q ∈ 0(C). Then, in rank i, AP = X(YC Q ±Y )RX for suitable Q ± ∈ 0(C),andV ≡ Y−1RY ∈ 0(C) by Lemmas 3.23 and 3.7. Hence, in rank i, AP = ±1 −1 (XY)C (Q ±V)(XY) ,whereQ ±V ∈ 0(C). Therefore the relation ∼i is transitive.

Lemma 3.25. —EverywordX is a conjugate in rank i ≥ 0 either of a 0-word or of awordAlP, where |A|≤|X|, A is either period of rank j ≤ i or a simple in rank i word and P represents in rank 0 an element of the subgroup 0(A). Proof. — If the statement were false for X, then, in rank i, X = YAsRY−1 for some (Y-reduced)essentialwordA with |A| < |X| and word R ∈ 0(A). By the inductive hypothesis, we may assume that A = ZBtPZ−1 in rank i where B is either a period of rank j ≤ i or a simple in rank i word, and P ∈ 0(B). We have that AsR = Z(BtP)sZ−1R in rank i.ThewordZ−1RZ is equal in rank i to a word Q ∈ 0(BtP) = 0(B) by Lemma 3.23 and Lemma 3.11. Therefore AsR = Z((BtP)sQ )Z−1.Since0(B) is normalized by B (by deﬁnition), X is a conjugate in rank i of a word of the form BstS for some S ∈ 0(B), as desired. Lemma 3.26. —IfawordX has a ﬁnite order in rank i, then it is a conjugate in rank i either of a 0-word or of a word AlR, where A is a period of rank j ≤ i and R ∈ 0(A). 84 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Proof. — If the claim were false, then, by Lemma 3.25, since 0(A) is normalized by A, AlR = 1 in rank i for some simple in rank i word A, R ∈ 0(A) and |l| >0. Then Al = R−1 in rank i and β¯|l||A| < |R|=0 by Lemmas 3.38, 3.39 and Lem- ma 2.7. The statement is proved by contradiction.

Lemma 3.27. —IfA and B are simple in rank i words and A is equal in rank i to XBlRX−1 for some word X,andsomewordR ∈ 0(B), then l =±1 and |A|=|B|.

Proof. — When proving the ﬁrst assertion by contradiction, one can assume that l ≥ 2. By Lemma 3.39 and Lemma 2.7 for a g-reduced annular diagram for the conjugacy of A and BlR,wehave2β¯|B| < |A|, whence |B| < |A| contrary to the simplicity of the word A in rank i. Then we have P ≡ XRX−1 ∈ 0(A) in rank i by Lemma 3.7, Lemma 3.11, and B = X−1(AP−1)±1X. Hence |A|=|AP−1|≥|B| since |P|=0 by deﬁnition of 0(A) (see property (Z3)) and the simplicity of B in rank i.

Lemma 3.28. —IfwordsX and Y are conjugate in rank i, and X is not a conjugate in rank i of a 0-word, then there is a word Z, such that |Z|≤α(¯ |X|+|Y|) and ZYZ−1 is equal to X in rank i.

Proof. — The proof is analogous to the proof of Lemma 18.5 [30]. One may apply Lemma 2.16 (which is an analog of Lemma 17.1 [30]) since X is not a conjugate of a 0-word.

Lemma 3.29. —Let∆ be a narrow contiguity diagram between sections q1 and q2 of , −1 its contour. Assume that it is of rank 0 and has contour p1q1p2q2 where q1 and q2 are paths labeled by A-periodic words with a simple in rank i period A. Assume that |q1|, |q2|≥2|A|, and o1 and o2 are vertices of q1 and q2 which are in the same phase, |(q1)− − (o1)|≤|q1|/2, |(q2)+ − (o2)|≤|q2|/2, and φ(o1 − o2) is a conjugate in rank i of a 0-word. Then the vertices (q1)− and (q2)+ are in the same phase too.

Proof. — Without loss of generality, one may assume that (q2)+ = o2 since the vertices o1 and o2 can be simultaneously shifted along the paths q1 and q2. Every Y-band of ∆ is a 0-bond between q1 and q2 because the narrow diagram ∆ of rank 0 is deﬁned by 0-bonds and by Lemma 3.2, there are no bands connecting different Y-edges of q1 (of q2)astheperiodA is simple in rank 0. Hence |p1|=0, and the vertex o1 can be connected with a vertex o0 of q2 byapathoflength0.Provingby contradiction, one may suppose that |V | >0for the word V ≡ φ((q1)− − o1). Then

|V|=|V | >0for the word V ≡ φ(o2 − o0). Denote U ≡ φ(o0 − o1).Then|U|=0, and the word VU is a conjugate of a0-wordinranki by the hypothesis of the lemma. Lemma 3.20 provides us with a decomposition V ≡ V1V2, where V2UV1 is equal in rank 0 to a word S of length at most 3γ |V|. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 85

Consider the vertex o of the subpath o2 −o0 that deﬁnes the decomposition V1V2 of the label V of o2 − o. Then the word V2UV1 is readable in ∆ starting with o:the sufﬁx V1 can be read in the beginning of o1 − (q1)+ since o1 and o2 are in the same phase. The vertex o can be connected with a vertex o of q1 by a path of length 0 (see Figure 10).

o1 V V 1 q1 o p1 U o q2 o 2 V1 V2 o0

FIG. 10.

Thus, up to a factor of length 0, the word V2UV1 is equal to a subword of q1 of ¯ the length |V2|+|V1|=|V|, which is not equal in rank i to a word of length ≤ β|V|, by Lemma 3.39 and Lemma 2.7. The inequality β¯ |V|≤3γ |V| contradicts the Lowest Parameter Principle. This completes the proof of the lemma.

The following Lemmas 3.30–3.32 are analogous to Lemmas 18.6–18.8 [30]. For agivenranki ≥ 1/2, they are proved by simultaneous induction on the sum L of lengths of periods.

Lemma 3.30. —Let∆ be a g-reduced diagram of rank i with contour p1q1p2q2,where −1 φ(q1) and φ(q2) are periodic words with a simple in rank i period A. Then the paths q1 and q2 ∆, | |, | | α| | | |, | | ( 5 + )| |. are A-compatible in provided p1 p2 < A and q1 q2 > 6 h 1 A (The inductive parameter L is equal to |A|+|A|.)

Proof. — The same argument, as in the proof of Lemma 18.6 [30] allows us to 5 ( )+ ( )− | |, | | | | assume that the vertices p1 and p1 are in the same phase, q1 q2 > 6 h A and |p1| < (α + 1/2)|A|=α¯|A|. In view of Lemma 3.7, to prove the lemma, one may −1 replace period A by a cyclic permutation. Therefore we suppose that both q1 and q2 start with the prefix A. Suppose that the word P1 ≡ φ(p1) is not a conjugate in rank i of a 0-word. In this case the annular diagram ∆¯ obtained after identification of the the long subpaths of q1 and q2 with equal labels, has no paths of length 0 surrounding the hole. There- fore we can apply Lemma 2.16 (which is Lemma 17.1 from [30]). The rest of the proof of Lemma 18.6 from [30] carries out with one minor change: by Lemma 3.25, m −1 the word D ≡ φ(p1) is equal in rank i to the product YB RY , R ∈ 0(B) instead of simply YBmY−1.ThewordDl is equal in rank i to YBmlQY−1, for some 0-word Q ∈ 0(B) since 0(B) is normalized by B. Then we have to define Z ≡ QY−1φ(s ), 86 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR instead of Z ≡ Y−1φ(s ) in [30]. This does not effect on the further proof because |Q l|=0 by the definition of 0(B). Assume now that the word P is a conjugate in rank i of a 0-word. Since q1 and −1 0 q2 are long enough (h = δ ), we can excise a narrow contiguity subdiagram ∆ of ∆ with contour p1q1p2q2 according to Lemma 2.11 (which can be applied because q1 and q2 are smooth sections of rank |A| by Lemma 3.39 in the smaller rank). Moreover by | |, | | 1 | | | |, | | α| | Lemma 2.11, we may suppose that q1 q2 > 2 h A and p1 p2 < A .Again,by

Lemma 2.11, there are vertices o1 and o2 of q1 and q2 respectively, such that the paths (p1)+ − o1 and (p1)− − o2 have the same label (say, X)and

−1 h |(q )− − o |, |(q )+ − o | < γ (α¯ + α)|A| < |A|. 1 1 2 2 6 −1 −1 Then the label of any path o1 − o2 is equal in rank i to X P X, and therefore it is (∆0) = , ( ) a conjugate in rank i of a 0-word. By Lemma 3.22, r 0 and the vertices q1 − ( ) and q2 + are in the same phase by Lemma 3.29 (see Figure 11).

! ! "" ! "" ! " ! " ! "" ! " ! o1 q1 " ! p ∆0 p p1 #1 !2 p2 # o ! 2 q2 ! # ! # ∆ ! # ! # ! # ! ## !

FIG. 11.

Every Y-band starting on q1 must end on q2 by Lemma 3.2. Hence one may φ( ) ≡ φ( )−1 ≡ . −1φ( ) = φ( )−1 suppose that q1 q2 U for some word U Then U p1 U p2 ( ) | | h | | | | −4φ( ) 4 ∈ ( ) in G 0 , and since U > 2 A >4A ,wehaveA1 p1 A1 0 A1 by Lemma 3.6, φ( ). where A1 is the cyclic shift of A which is the preﬁx of q1 Thus the paths q1 and q2 are compatible by (Z3.1) since A is simple in rank 1/2 (recall that i ≥ 1/2). Therefore the paths q1, q2 are also compatible by Lemma 3.7, and the lemma is proved.

m1 m2 Lemma 3.31. —LetZ1A Z2 = A in rank i, m = min(m1, m2), and A be a simple word in rank i. Then each of the words Z1, Z2 is equal in rank i to a product of a power of A and ( ) | |+| | (γ( − 5 − )− )| |. a word representing an element of the subgroup 0 A provided Z1 Z2 < m 6 h 1 1 A (The inductive parameter L is equal to |A|+|A|.)

Proof. — As in Lemma 18.7 [30] we can consider a g-reduced disc diagram ∆ − , φ( ) ≡ m1 φ( 1) ≡ m2 φ( ) = of rank i with the contour p1q1p2q2 where q1 A , q2 A , p1 Z1, NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 87

φ(p2) = Z2. By Lemmas 3.38, 3.39, q1 and q2 are smooth of rank |A|.ByLem- ∆ ∆ mas 3.38 and 2.15, we can excise from a subdiagram with contour p1q1p2q2 where | |, | | α| |, | | | |−γ −1(| |+| |+| |) p1 p2 < A qj > qj Z1 Z2 A > | |−( − 5 − )| |=( 5 + )| |, = , . m A m 6 h 1 A 6 h 1 A j 1 2 ∆ ∈ By Lemma 3.30, q1 and q2 arecompatiblein . So there exist phase vertices o1 q1 and o2 ∈ q2 such that the label T = φ(t) of a path t connecting these vertices represents an element of 0(A) in rank i.Thenφ((q1)− − o1) and φ((q2)+ − o2) are powers s1 s2 of A,wehaveanequalityofthefollowingforminranki: Z1 = A TA .Since0(A) is normalized by A in rank 0, the lemma is proved.

Lemma 3.32. —Let∆ be a g-reduced disc diagram of rank i with the contour p1q1p2q2, where φ(q1) and φ(q2) are periodic words with simple in rank i periods A and B, respectively, and | |≥| |. | |, | | α| |, | | 3 | | | | | |. A B Assume that p1 p2 < B q1 > 4 h A and q2 >hB Then the word A is a conjugate in rank i of a product B±1R, where R represents an element of the subgroup 0(B), and −1 −1 |A|=|B|. Moreover, if the words φ(q1) and φ(q2) start with A and B , respectively, then A −1 ±1 is equal in rank i to φ(p1) B Rφ(p1). (The inductive parameter L is equal to |A|+|B|.) Proof. — The proof is similar to the proof of Lemma 18.8 [30], but, at the very s s end, by Lemma 3.31, we obtain the equality Z1 = B R (instead of Z1 = B )with R ∈ 0(B).Thens =±1 and |A|=|B| by Lemma 3.27.

Lemma 3.33. —Let∆ be a disc diagram of rank i with the contour p1q1p2q2, where both φ(q1) and φ(q2) are periodic words with a simple in rank i period A and |p1|, |p2| < α|A|. Then |q1|, |q2|≤h|A|. Proof. — We prove the statement by contradiction. The proof is parallel to the proof of Lemma 18.9 [30] up to the application of Lemma 3.32 (instead of Lem- ma 18.8). This lemma gives the equality AR = Z−1(A−1)l Z in rank i,wherel =±1, R ∈ 0(A), |Z| < (1 + α)|A|. Then, in case l =−1, Z−1AkZ = AkQ in rank i, k ¯ ¯ where Q ∈ 0(A), because of (Z3) where φ(q1) ≡ A A, |A| < |A|. The rest of the argument of the proof of Lemma 18.9 from [30] in this case remains valid because 0(A) is normalized by A in rank 0 and all elements in 0(A) have length 0 (so the corresponding factors do not affect length estimates). In case l = 1, the equality AR = Z−1A−1Z,inranki, implies An = Z−1A−nZ by Lemma 3.12. Hence Z2 commutes with An in rank i. Therefore Z2 commutes with any Asn, s = 1, 2,....Takings sufﬁciently large, we can apply Lemma 3.31 and conclude that Z2 = Ad P in rank i, with P ∈ 0(A). Then again, Z2n = Adn in rank i by Lem- ma 3.12. Therefore the word Z commutes with Adn in rank i. But the equality An = Z−1A−nZ implies that Z−1AdnZ = A−dn in rank i. Hence A2dn = 1 in rank i. This would contradict the deﬁnition of a simple word and Lemma 3.26, if d = 0. 88 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Consequently d = 0 and Z2 = P in rank i. Then Z has finite order in rank i by property (Z1.2). This order must divide n by property (Z1.2), Lemma 3.12 and Lemma 3.26 since Bn = 1 in rank i for every period B of rank ≤ i. Therefore Z2 = P ∈ 0(A) implies that Z ∈ 0(A) in rank i as n is an odd number. By Lemma 3.13 Z commutes with An in rank i. It follows that A2n = 1 in rank i. This is the final contradiction, as simple in rank i word A has infinite order in rank i by Lemma 3.26.

Lemma 3.34. —Let∆ be a g-reduced disc diagram of rank i with the contour p1q1p2q2, where φ(q1) and φ(q2) are periodic words with simple in rank i periods A and B. Assume that | |, | | α| |, | | ( + γ )| |, | | 1 ε | |. p1 p2 < B q1 > 1 2 A q2 > 2 n B Then the word A is a conjugate in rank i ±1 −1 ±1 of a word B R, where R ∈ 0(B). Moreover it is equal in rank i to φ(p1) B Rφ(p1) if the −1 −1 word φ(q1) starts with A and φ(q2) starts with B .

Proof. — To prove the lemma, one should repeat the argument from Lemma 19.1 [30] up to the reference to Lemma 3.31, instead of Lemma 18.7 [30] which provides us with the additional factor R as compared with [30].

Lemma 3.35. —Let∆ be a g-reduced disc diagram of rank i with the contour p1q1p2q2, where φ(q1) and φ(q2) are periodic words with simple in rank i periods A and B;|q2|≥εn|B|, |p1|, |p2| < ζnc for c = min(|A|, |B|). Then either |q1| < (1 + γ)|A|, or A is a conjugate in rank i of a word B±1R where R ∈ 0(B). Furthermore, equality A ≡ B±1 implies A ≡ B−1 and the paths q1, q2 are A-compatible in ∆.

Proof. — The statement can be reduced to the claim of Lemma 3.34 in the similar way as in Lemma 19.2 [30].

Recall that i+ means 1 if i = 1/2,andi + 1 if i ≥ 1.

n Lemma 3.36. —1)If|X|≤i+ then X = 1 in rank i+. 2) period A of rank i+ is simple in any rank j ≤ i. 3) Let A and B be periods of ranks k, l ≤ i+,andA ∼j B for some j

Proof. — 1) If X is a conjugate of a word of length 0 in rank i+, the statement follows from (Z1.2). Otherwise by Lemma 3.25 X is a conjugate in rank i of a word of the form AlP with |A|≤|X|, P ∈ 0(A) and A is either a period of rank ≤ i or a simple l n ln ln word in rank i.Then(A P) = A by Lemma 3.12. If |A|

2) Follows from the inclusion of the set of relators of rank j into the set of relators of rank i for j ≤ i. ±1 −1 3) If k were less than l,forexample,thenB = XA PX in rank l− for some X and P ∈ 0(A) as it was shown at Lemma 3.24, but this contradicts the choice of periods of rank l.Ifk = l,thenA ≡ B by the deﬁnition of the set Xl.

Lemmas 3.37–3.39 will be proved by simultaneous induction on the type of diagrams.

Lemma 3.37. —Letπ beahubinag-reduceddiagram∆ of rank i+. Then (1) the degree of contiguity of arbitrary cell Π to π is less than ε; (2)arbitrary contiguity subdiagram Γ of π to a cell Π or to a section p of the boundary ∂(∆), labeled by a A-periodic word where A is a simple in rank i+ word or a period of rank ≤ i+, is of rank 0, provided in the last case that there are no cells compatible with p, in ∆.

Proof. — Consider a contiguity subdiagram Γ between a hub π and another cell Π. Let E be one of the bonds deﬁning Γ. If r(E) >0, then the principal cell of E has the contiguity degree to π at least ε, and this contradicts the statement of the lemma for a subdiagram of a smaller type (the subdiagram does not contain Π). Thus E is a0-bond. Furthermore the same argument shows that a cell π of positive rank from Γ possesses no contiguity subdiagram to π (to Π if Π is also a hub), with degree ≥ ε.If Π is a cell of rank ≥ 1, then by Lemmas 3.38, 3.39 for Γ and Lemma 3.17, one may apply Lemma 2.3 and conclude that a contiguity degree of π to Π is less than α¯. SincesidearcsofΓ are of length 0, this contradicts Lemma 2.4 because ε + α¯ < γ.¯ Thus r(Γ) = 0 if π = Π. If Π is a hub, then (Π, Γ,π) < ε by property (Z2.3) because otherwise a subdiagram containing hubs π and Π could be replaced by a subdiagram of rank 0, but ∆ is a g-reduced diagram. If Π has rank ≥ 1, then every maximal Y-band of Γ, starting on Π, must end on π by Lemma 3.2, and so Γ Π cannot contain a subword of length > |A| since every letters of φ(Γ π) are distinct by (Z2. 2). Therefore (Π, Γ,π) ≤ ι < ε by A1, and the lemma claim is proved for a contiguity subdiagram between a hub and another cell. If π = Π,thenagainthebondE has rank 0, because otherwise the principal cell π of this bond differs from π,and,bypreviouscase,π possesses no contiguity subdiagram of degree ≥ ε to π contrary to the deﬁnition of a bond. Then r(Γ) = 0 as above, and the assumption (π, Γ,π) > ε would contradict property (Z2.2). Finally, if Γ is the contiguity subdiagram of the hub π to the boundary section p given in the lemma hypothesis, then again the rank of the bond E is 0 as above, and then the same argument shows that r(Γ) = 0.

Lemma 3.38. — A g-reduced diagram ∆ of rank i+ is an A-map. 90 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Proof. — 1) Hubs have rank 1/2 and perimeter at least n by property (Z2.1). The perimeter of a cell with a period of a positive integer rank j ≤ i+ is equal to nj. Thus ∆ satisﬁes condition A1. 2) Let q be a subpath of ∂(Π),wherer(Π) = j ≤ i+, and let |q|≤max( j, 2). Let p be a path homotopic to q in ∆, and Γ denote the subdiagram with contour p−1q. If Π does not occur in Γ, then the type of Γ is smaller than that of ∆,andΓ is an A-map by the inductive hypothesis. If also |q|≤2 and |p| <2,then|∂(Γ)|≤3, and r(Γ) = 0 by Lemma 2.8. Then φ(p) = φ(q) in G(0), which is impossible by the cyclic irreducibility of periods and, if Π is a hub, by (Z2.2). We suppose next that |q|≤j and j>1. If |p| < |q|, then |∂(Γ)| <2j, and by Lemma 2.8, the perimeters of all cells of positive rank in Γ are less than 2jβ¯ −1, and so r(Γ) <2jβ¯ −1n−1

|q|≤2(n − 2)−1|q |≤2β¯ −1(n − 2)−1|p| < |p|.

It remains to suppose that Π is a hub. Then, by Lemma 3.37, there is no cell in Γ with a contiguity subdiagram to q of degree ≥ ε. Hence q is a smooth section in ∂(Γ ).Thenagain|q|≤2(n − 2)−1β¯ −1|p| < |p|, which completes the veriﬁcation of condition A2. 3) Let Γ be a contiguity subdiagram of a cell of rank j to a cell of rank k in ∆, with p1q1p2q2 = ∂(π, Γ, Π) and (π, Γ, Π) ≥ ε. Then k ≥ 1 by Lemma 3.37, and Γ is an A-map by the inductive hypothesis. Assume ﬁrst that j = 1/2. Then r(Γ) = 0 by Lemma 3.37. By Lemma 3.2 every maximal Y-band of Γ starting on q2 must end on q1, and therefore, by property (Z2.2), all the edges of q2 have distinct labels. Hence |q2|≤|A|=k<(1 + γ)k where A is the period of Π.

Then suppose that j ≥ 1. In this case, q1 and q2 are smooth sections of ranks j and k, respectively, in ∂(Γ) by Lemma 3.17 2) and Lemma 3.39 applied to Γ. Thus, applying Lemma 2.1 to Γ, |p1|, |p2| < ζn min( j, k). Hence, by Lemma 2.14, r = r(Γ) < min( j, k), and, by Lemma 3.36, the periods B, A corresponding to π and Π are simple in rank r. Thus, applying Lemma 3.35 to Γ,either|q2| < (1 + γ)|A|= (1 + γ)k (that is, condition A3 holds), or A = XB±1RX−1 in rank r for R ∈ 0(B),and NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 91 therefore A ≡ B by Lemma 3.36. Hence, by Lemma 3.35, q1 and q2 are compatible, contradicting Lemma 3.17 1). This completes the proof.

Lemma 3.39. —Letp be a section of a contour of a g-reduced diagram ∆ of rank i+ and one of the following two conditions holds. (1) The path p is not a cyclic section and φ(p) is an A-periodic word where A is a period of a rank k ≤ i+, and there are no cells in ∆, compatible with p, or a simple word in rank i+. m (2) The path p is a cyclic section and φ(p) ≡ A P where A is a simple in rank i+ word and P ∈ 0(A). Then p is a smooth section of rank |A| in ∂(∆).

Proof. — Case 1. Suppose ﬁrst that assumption (1) from the lemma holds. Let us check two conditions from the deﬁnition of a smooth section. 1) Let q be a subpath in p, |q|≤max(|A|, 2),andt a path homotopic to q in ∆. The subdiagram Γ with contour qt−1 is an A-map by Lemma 3.38. If |t| < |q|≤2, then |∂(Γ)|≤3, which gives a contradiction as in the proof of Lemma 3.38. We can thus assume that |q|≤|A|. If A is a simple in rank i+, word then it is not a conjugate in rank i+ of a shorter word, and so we have, for the subword φ(q) of a cyclic permutation of A, |φ(q)|≤|φ(t)|, since φ(q) = φ(t) in rank i+. This enables us to assume that r(A) = k ≤ i+. If we now suppose that |t| < |q|, then |∂(Γ)| <2k, and by Lemma 2.8, r(Γ) = j

Case 2. Suppose now the assumption (2) of the lemma holds. The label of the contiguity arc q2 in item 2) (and φ(q) in item 1)) can be of the form U1PU2 where U1U2 is a phase decomposition of an A-periodic word. By the deﬁnition of 0(A),by 92 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Lemma 3.6 and by Lemma 3.7 PU2 = U2P in G(0),whereP ∈ 0(A ), |P |=0,

A is a cyclic permutation of A and U ≡ U1U2 is an A -periodic word with phase decomposition U · 1. Hence one can transform Γ by adding a subdiagram of rank 0 −1 whose contour labelled by UP (U1PU2) . The new diagram Γ can be considered as in Case 1, because |P |=0 and P can be can be considered a part of the label of thesidearcofΓ . The lemma is proved. This together with Lemma 3.36 1) complete the proof of Proposition 3.19. Proposition 3.19 immediately implies the following statement that shows that one can construct analogs of HNN extensions in the class of groups of odd exponent n 1. Corollary 3.40. —LetG =A | R be a group satisfying the identity xn = 1 for some odd n 1.LetPi, Q i be subgroups of G, i = 1, 2,...,k, k ≥ 1,andforeachi = 1,..,k,let φi : Pi → Q i be an isomorphism. Let H be the (multiple) HNN extension of G corresponding to these pairs of isomorphic subgroups, = , , ..., | , −1 = φ ( ), ∈ , = ,..., . H A t1 tk R tiuti i u for all u Pi i 1 k

Consider A as a set of 0-generators and Y ={t1,...,tk} as the set of non-zero generators of H.Sup- pose that H satisfies the conditions (Z3.1) and (Z3.2) (or (Z3.1) and (Z3.2’)). Then the natural homomorphism G → H/Hn is an embedding. Proof. — Indeed, notice that H obviously satisfies conditions (Z1). Conditions (Z2) also are satisfied because there are no hubs. Conditions (Z3) are satisfied by our assumption. Hence by Proposition 3.19 every g-reduced diagram over some graded n presentation R∪S1 ∪S2 ∪... of H/H is an A-map, and every relator in Si, i = 1, 2,... is of the form An where A is a cyclically reduced word of Y-length i. Let W be a word in A which is equal to 1 in H/Hn.Let∆ be a g-reduced van Kampen diagram with boundary label W. If the rank of ∆ is at least 1 then by Lemma 2.5 ∆ contains a cell Π of rank ≥ 1 and a contiguity subdiagram of rank 0 of Π to ∂(∆) with non-zero contiguity degree. This implies that the label of ∂(∆) must contain a letter from Y, a contradiction (we assumed that W contains only letters from A). Hence ∆ has rank 0, so ∆ is a diagram over R.HenceW = 1 in G. Remark 3.41. — Sergei Ivanov also proved Corollary 3.40 [17]. We added this corollary in our paper after the second author attended Ivanov’s talk at the Geometric Group Theory conference in Montreal (May, 2002). A particular case of that corollary has been obtained by K.V. Mikhajlovskii [24]. He considered the case when k = 1, −1 each of the subgroups P1 and Q 1 is malnormal (i.e. P1 ∩ xP1x ={1} if x ∈ G\P1, −1 −1 Q 1 ∩ xQ 1x ={1} if x ∈ G\Q 1), and xP1x ∩ Q 1 ={1} for every x ∈ G.Inthat case 0(h) ={1} for every non-zero cyclically minimal element h ∈ H. Hence condition (Z3) trivially holds. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 93

4. Subgroups of the free Burnside groups

In this section we prove certain properties of free Burnside groups. Although Theorem 4.4 below is of independent interest, this section plays an auxiliary role in the paper and can be skipped in the ﬁrst reading.

4.1. Sets of words with small cancellation

Let A ={a1, ..., am} be an alphabet. For a constant c>1, we say that a word v is c-aperiodic if for every word A,therearenonon-emptyA-periodic subwords w of v with |w|≥c|A|.

Lemma 4.1. —Foreveryε >0, one can ﬁnd m = m(ε) such that the set of (positive) (1 + ε)-aperiodic words in the alphabet Am ={a1, ..., am} is inﬁnite. Moreover, the number f (i) of such words of length i ≥ 1 is greater than m(1−ε/2)i.

Proof. — First notice that one can assume that ε is sufficiently small and fix a number m such that ε ε ε + 1 −1+ + 2 1− 1 (14) m − m 2 1+ε − m 2 1+ε >m 2 − m 1+ε . Obviously, f (1) = m>m1−ε/2. We will prove the inequality f (i +1) >m1−ε/2f (i), for i ≥ 1, by induction on i. By adding a letter to a (1 + ε)-aperiodic word w of length i on the right, one obtains mf (i) words of length i + 1. But some of these words w may contain non- empty A-periodic subwords v with |v|≥(1 + ε)|A| for some |A|. We need an upper estimate for the number of such “bad” possibilities. Clearly, the subword v must be a suffix of w, w ≡ uv, since otherwise v would be a subword of the prefix w of length i in w. Further we may assume that the word v starts with the period A, and the previous argument shows that A is uniquely determined by v, because otherwise word w were not (1 + ε)-aperiodic. Since |v|≤ (1 + ε)|A|, the total number of the words v of length k, which can occur, is at most k m 1+ε . Therefore the number of “bad” products w ≡ uv with |v|=k and |u|=n+1−k, k is at most f (n + 1 − k)m 1+ε . By using the inductive hypothesis, one can substitute the first factor by f (n)m(1−k)(1−ε/2). Then, by summing over k ≥ 2, we obtain ∞ (1−ε/2)(1−k)+ k f (n + 1) >mf(n) − f (n) m 1+ε = k=2 − + ε + 2 1 2 1+ε m 1−ε/2 f (n) m − ε >m f (n) −1+ + 1 1 − m 2 1+ε by (14), as required. 94 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Lemma 4.2. —Foranyε >0, there exists m = m(ε) and an infinite set S of (positive) words A1, A2, ... in the alphabet Am ={a1,...,am} such that: (1) every reduced product, whose factors belong to S, has no non-empty A-periodic subwords of length ≥ (1 + ε)|A| unless the word A is freely conjugate to a product of some words of S; (2) suppose that A ≡ UV and A ≡ UV ,orA ≡ V U and A ≡ V U, are distinct cyclic permutations of the words of S (that is, U is a common prefix or a common suffix). | | ε (| |, | |). Then U < 10 min A A (3) |Ai|≥n, i = 1, 2,....

Proof. — We can assume that ε <1/3 and choose a number m1 = m1(ε1) for ε = ε/ , , ... A 1 10 and an infinite set of positive words B1 B2 in alphabet m1 , according = ≥ ε−1. to Lemma 4.1. Then define m sm1 for a number s 2 1 Set Ai = Bi,1...Bi,s for i = 1, 2,...,whereBi, j are the copies of the word Bi in the A ={ ,..., } { , , ...} alphabet j a1j am1 j . Throwing out, if necessary, a finite subset of A1 A2 we can assume that |Ai|≥n. Hence property (3) holds. Property (2) follows from the ≥ ε−1 A inequality s 2 1 because the alphabets j are disjoint. −1, ε In particular, in any product AiAj we can cancel at most 1 (in length) of Ai = , −1 or Aj if i j and no letter of Bi,s−1 or Bj,s−1 can be cancelled in the product. Similar −1 . remark concerns the product Ai Aj To prove (1), suppose that v is a non-empty A-periodic subword of a reduced ≡ ¯ ±1... ¯ ±1 ±1... ±1, | |≥( + ε)| |. form w Ai1 Ail of a product Ai1 Ail and v 1 A We may assume that v ≡ A A A , where A ≡ A A and |A |≥ε|A|, because otherwise we can just shorten the subword v. ¯ ±1 Assume first that both occurrences A arecontainedinthesamefactorAir of w. ±1 ... Then v is a subword of the product Bir 1 Birs, and the both occurrences of A are . , present in some Bir j Hence v is a subword of Bir j contrary to the choice of the words B1, B2.... ¯ ±1 Now assume, the first A of the decomposition of v starts in some Ail ,andthe ¯ ±1 . last A ends in some Air with r>l Then the definition of the words Ai implies that s−3 2 ≤ ≤ . v must contain at least 2 > 5 s factors Bil j of some Aik where l k r Then 2 |A |≥ε(1 + ε)−1|v| > ε(1 + ε)−1 |A | >3ε |A |. 5 ik 1 ik ( )±1 . By the choice of s, A contains a subword Bik,tBik,t+1Bik,t+2Bik,t+3 for some t We will suppose that the exponent is equal to +1. Hence a cyclic permutation of A is ( )...( ). freely equal to a word Bik,t+2Bik,t+3 Bik,tBik,t+1 This subword of w must be freely ±1 equal to a cyclic permutation of a product of a number of factors Ai , because there is only one occurrence of the form Bj,qBj,q+1 in all the words A1, A2, .... The lemma is proved. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 95

Remark 4.3. — In the proof of Theorem 1.1, we do not need infinite collections of words satisfying the conditions of Lemma 4.2. We only need a collection of words satisfying that conditions and consisting of Cm words where C is a constant 20 depending on n (but not on m), C>n> ε ,andm is a sufficiently large integer. An explicit and easy construction of such collections of words can be found in [35]. Let us repeat this construction here. Let us fix integers r>C2 and m = rn. Consider an r2 × n-matrix M where every odd-numbered column is equal to (1, 2, ..., r, 1, 2, ..., r,...,1, 2, ..., r) and every even-numbered column is equal to (1, 1, ..., 1, 2, 2,...,2, ..., r, r, ..., r).

Now let us replace every number i in column j of M by letter ai, j.Rowsoftheresult- ing matrix can be viewed as words of length n over the alphabet with m = rn letters {ai, j, i = 1,...,r, j = 1, ..., n}. For example the first word is equal to a1,1a1,2...a1,n,etc. Let S ={w1, ..., wr2 } be the collection of these words. Let us call words from S and their inverses blocks. It is mentioned in [35] (and obvious) that no two different words in S have a common subword of length 2. Condition (3) of Lemma 4.2 holds by definition of 20 the blocks. Since n> ε , condition (2) of Lemma 4.2 holds. = s1 ... sk To prove (1), consider a reduced product W wi1 wik of words from S, =± sj sj+1 = = , ..., − si 1,wherewij wij+1 1, j 1 k 1. Suppose that W contains an A-periodic subword V where |V| > (1 + ε)|A|.Sincen>4/ε, we can assume that V contains a subword AB where B is the beginning of A of length 4. Then either the first two letters of B are in one block of W or the last two letters of B are in one block of W. Since every block is completely determined by any of its 2-letter subwords, we can shift A to the left or to the right in W,replacingA by a cyclic shift A of A,suchthat A starts with the beginning of a block and ends with the end of a block. Then A is a product of blocks as required.

4.2. Subgroups of B(m, n) satisfying the congruence extension property Recall that a subgroup H ≤ G satisﬁes the congruence extension property if for every normal in H subgroup N there is a normal subgroup L of G,suchthatN = L ∩ H. In this section, we prove

Theorem 4.4. — For any sufficiently large odd n, there exists m = m(n) such that the free Burnside group B(m, n) has a subgroup H with three properties: (1) H is isomorphic to the free Burnside group B(∞, n) of the infinite countable rank; (2) H satisfies the congruence extension property in B(m, n). 96 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

(3) the words A1, A2,... representing the free generators of H can be effectively chosen in accordance with Lemma 4.2, and for every l ≥ 1, the free Burnside subgroup generated by A1, ..., Al , also possesses the congruence extension property in B(m, n).

The statements (1) and (2) constitute Theorem 1.2. The proof of Theorem 4.4 follows very closely Section 6 of [30] although we repeat a few notation and definitions from [30]. In particular, we use the positive parameters α, β, γ, δ, ε, ζ, ι, n from Section 2. Let m = m(ε) be taken according to Lemma 4.2. Then we fix the infinite set of words A1, A2, ... which exists by Lemma 4.2. By Lemma 4.2 (3) the words Ak are sufficiently long, say, |Ak|≥n. Let H be the subgroup generated by all words A1, A2,... in the free Burnside group B(m, n) with the free basis A ={a1,..,am}. Suppose that v is a (cyclically) reduced word of length t in certain variables η η , ..., ( , ..., ) ≡ 1 ... t x1 xs. The (cyclic) word v A1 As Ai1 Ait obtained by replacing Ai for xi will be called a (cyclic) H-word and will be considered together with its decomposition η η 1 , ..., t . into factors Ai1 Ait

Lemma 4.5. —ThewordsA1, A2..., satisfying conditions (1) and (2) of Lemma 4.2, freely generate a free Burnside subgroup (of infinite countable rank) of exponent n in the group B(m,n). Proof. — To prove the lemma, we consider the graded presentation of B(m, n) from Section 18 in [30]. Assume there is a non-trivial relation v(A1,...,Al) = 1.Then there is a g-reduced diagram ∆ of some rank i>0whose boundary label φ(q) is freely equal to an H-word. Proving by contradiction, we may suppose that the number of cells of ∆ is minimal. By Lemma 2.5, there is a cell Π of ∆ and a contiguity subdiagram Γ with (Π, Γ, q) ≥ ε. Since nε >2, we obtain, by Lemma 4.2(1), (3) that Π corresponds to a period A where A is freely conjugate to an H-word A¯ . By Lemma 4.2(2) the boundary label φ(q ) of the subdiagram ∆ , obtained from ∆ by cutting off the cell Π, is freely equal to an H-word also, and φ(q) is freely equal to the product of φ(q ) and a word XA¯ nX−1 for some H-word X. By the minimality of ∆, diagram ∆ corresponds to a relation v (A1, ..., Al) = 1 such that v (x1, ..., xl ) follows from the Burnside n identity x = 1. Then the relation v(A1, ..., Al) = 1 follows from Burnside relations too. Let ∆ be a map. A smooth section q of rank 1 in the contour ∂(∆) is said to be ε-section if for any cell π of ∆ and any contiguity subdiagram Γ, (π, Γ, q) < ε. Let us now fix an arbitrary normal subgroup N in H. We would like to define a group presentation of the factor group of B(m, n) by the normal closure of N in B(m, n). We denote by T the set of all H-words (in the alphabet a1, ..., am)representing elements of N, that are cyclically reduced as elements of the free subgroup H (of the NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 97 absolutely free group) with generators A1, A2, ....Inparticular,T contains all powers n of the form v(A1,...,As) . The following alteration of the definitions from Section 18 of [30] depends on N. We define G(0) = F(A ) to be the absolutely free group with the empty set R0 of defining words. For every i ≥ 0 anon-trivialwordA is called simple in rank i if it is not conjugate in rank i (i.e. in group G(i)) either to a power of a period B of rank k, where 1 ≤ k ≤ i, or to a power of a word C,where|C| < |A|,ortoanH-word. Every word of T is included in the system R1 of relators of rank 1. (Let us call n, ..., n R . them T-relators.) We also include words a1 am in the system 1 Every word ak of length 1 is, by definition, a period of rank 1, and G(1) =A | R1. For ranks i ≥ 2, the definitions of periods of rank i and the group G(i) = A |Ri, G(∞), given in Section 18 of [30] remain valid. Instead of Lemma 18.1 [30] we claim now that the following statement holds. Lemma 4.6. —EverywordX over A is a conjugate in rank i ≥ 0 either of a power of some period of rank j ≤ i or of a power of a simple in rank i word, or of an H-word. The proof can be easily derived from the definition as in [30]. Analogs of Lemma 18.2, Corollaries 18.1 and 18.2 are not used in the proof, so we do not need them now. The following statement replaces Lemma 18.3 [30]. Lemma 4.7. —IfawordX has a finite order in rank i, then it is a conjugate in rank i of a power of a period of rank k ≤ i or of an H-word. There are no changes in the proof as compared with [30]. Assume that Π is a T-cell (i.e. it corresponds to a T-relator) in a g-reduced annular diagram ∆ of rank i, and Γ is a contiguity diagram of the cell Π to itself, such that the hole of ∆ is “surrounded” by Π and Γ, i.e. the union of Π and Γ does not belong to any disc subdiagram of ∆ (see Figure 12). The subdiagram Γ is of rank 0 by Lemma 4.13 (1) applied for a smaller rank. If the degree of Γ-contiguity of Π to Π is at least ε, we call Π a hoop. The reduced ¯−1 ¯ boundary q of a hoop can be decomposed as q1pq2p ,wherep, p are the contiguity arcs of Γ, φ(p¯) is freely equal to φ(p),andso|p|, |p¯|≥ε|q|.Inviewofproperty (2) of Lemma 4.2, there is a subpath p of p,suchthatφ(p ) is a unique subword of ±1. § a unique word Ak (More details of such an argument can be found in 2 of [31].) This implies that the words φ(q1) and φ(q2) are freely conjugate to some H-words. −1 −1 Besides, it follows from T-relations that Xφ(q1)X = φ(q2) for some H-word X. Lemma 18.4 in [30] remains unchanged: Lemma 4.8. —IfA and B are simple in rank i and A is equal in rank i to XBlX−1 for some X then l =±1. 98 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

∆ p¯ p

q1 q2

FIG. 12.

The proof does not change much also. Notice only that by deﬁnition, a simple word of rank i ≥ 1 cannot be conjugate in rank i of an H-word. Therefore the diagram ∆ from the proof of Lemma 18.4 [30], cannot have hoops, and so the analog of Lemma 19.4 [30] (see Lemma 4.15 below) is applicable.

The formulation of Lemma 18.5 changes as follows.

Lemma 4.9. —IfwordsX and Y are conjugate in rank i and X is not a conjugate of any H-word in rank i then there exists a word Z such that X is equal to ZYZ−1 in rank i and |Z|≤α(¯ |X|+|Y|).

The additional assumption that X is not a conjugate of an H-word again allows us to apply Lemma 4.15, the analog of Lemma 19.4 [30].

Consider a section q of a contour of some diagram ∆,whereφ(q) is freely equal to an H-word. We call q an H-section. Assume that there is a T-cell Π in ∆ with boundary path p starting with a vertex o of ∂(Π), and there is a path t having no self-intersections, which connects o and a vertex o cutting q into a product q1q2. We −1 say that Π is compatible with q, if the boundary label of the path q1t ptq2, without self- intersections, is freely equal to an H-word. Obviously, one can cut such a cell out of ∆ replacing q by another H-section. The definition of compatibility can be generalized inthecasewhenφ(q) is a subword of a reduced form W of an H-word with |φ(q)|≥ ε|W| in view of Lemma 4.2(2). (Such a path q is also an H-section by definition.) Similarly, one defines the compatibility of two T-cells in a diagram. A pair of two compatible T-cells can be substituted by (at most) one T-cell. Therefore a g-reduced diagram of some rank i cannot contain pairs of compatible T-cells or j-pairs in the terminology of [30] for j = 1.ForBurnside cells (i.e. cells corresponding to relations of the form An = 1 where A is a period of rank j>0) we are using the same definition of j-pairs as in [30]. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 99

Lemma 4.10. — Assume that a word W is a conjugate of an H-word in rank i. Then W = UVU−1,inranki, for some H-word V and some word U, such that |V|≤β¯ −1|W| and |U|≤2|W|.

Proof. — Consider a g-reduced annular diagram ∆ for the conjugacy of W to some H-word V. We may assume that there are no hoops in ∆, because otherwise one can replace ∆ by a diagram of a smaller type, that also demonstrates the conjugacy of W to a subword of the boundary label of a hoop. For the same reason, we may assume that ∆ has no T-cells compatible with the contour, say p, of ∆ labeled by V. Then ∆ is an A-map by Lemma 4.15, and p is a smooth section of ∂(∆) by Lem- ma 4.16. Then |V|≤β¯ −1|W| by Lemma 2.7, and the word U can be taken in such a way that |U|≤α(¯ 1 + β¯ −1)|W|≤2|W| by Lemma 4.9.

The following Lemma implies malnormality of the subgroup H in B(m, n).

Lemma 4.11. —LetV and W be two non-trivial in rank iH-words and W = UVU−1 in rank i. Then the word U is itself equal in rank i to an H-word.

Proof. — Let ∆ be a g-reduced annular diagram of rank i with contours p and q labeled by V and W respectively. One can assume that there is a path t = q− − p− in ∆,suchthatφ(t) = U in rank i. Then it sufﬁces to show that φ(t) is freely equal to an H-word. One can suppose that ∆ has no T-cells compatible with the contours, because otherwise our problem reduces to a similar problem for an annular diagram of smaller type. If ∆ has a hoop, then t can be constructed as a product t1t2t3 where the path t2 cuts the hoop and t1, t3 are the cutting paths for annular subdiagrams of smaller types whose labels are H-words by induction. Thus, we may assume that there are no hoops in ∆. Then both p and q are ε-sections by Lemma 4.14. It follows that ∆ has rank 0, because otherwise the inequality γ¯ >2ε, Lemma 2.4 and Lemma 4.15 lead to a contradiction. Then the desired property of ∆ follows from the malnormality of the subgroup generated by the words A1, A2, ... in the absolutely free group F(A ), which, in turn, can be easily derived from the statement of Lemma 4.2(2).

Next as in [30, Section 18] we prove analogs of Lemmas 18.6, 18.7, 18.8 by a simultaneous induction on the sum L of the two periods involved in the formulations of these lemmas. The formulations of these lemmas do not change.

Lemma 4.12 (Analog of Lemma 18.6 [30]). — Let ∆ be a g-reduced disc diagram of φ( ) φ( −1) rank i with contour p1q1p2q2 where q1 and q2 are periodic words with period A simple in 100 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR | |, | | α| | | |, | | ( 5 + )| | ∆ rank i.If p1 p2 < A and q1 q2 > 6 h 1 A ,thenq1 and q2 are A-compatible in . (The inductive parameter is L =|A|+|A|.)

Proof. — As in the proof of Lemma 18.6 from [30], assume that A is a preﬁx φ( ) φ( −1) / | | of both q1 and q2 by making if necessary p1 a little longer (by at most 1 2 A ). The new path is also denoted by p1. Notice ﬁrst that if φ(p1) is not a conjugate of an H-word in rank i then in any g-reduced annular diagrams of rank ≤ i where one of the contours is labeled by φ(p1), there are no hoops. So the proof proceeds as in [30]. Otherwise, as in [30], consider the annular diagram ∆¯ of rank i with boundary −1 labels W ≡ φ(p1) and W ≡ φ(q )φ(p2) .ThenW, W are conjugate in rank i of some H-word. If W = 1 in rank i then there is nothing to prove. Thus we assume that the word W is non-trivial in rank i. Inequalities (2), (3) from Section 18 of [30] are still valid:

|W| < α¯|A|, 5 |W | < (β¯ −1 − 1) δ−1|A|+|A| + 2|A| < (β¯ −1 − 1)(|q |+2|A|). 6 2 By Lemma 2.10, one can ﬁnd H-words V and V such that

|V|≤β¯ −1|W| < β¯ −1α¯|A| < |A|, 5 |V | < β¯ −1(β¯ −1 − 1) δ−1|A|+|A| < βδ−1|A|, 6

− and, in rank i, W = UVU−1, W = U V U 1, where |U| <2|A|, |U | <2βδ−1|A| by Lemma 4.10. −1 = , ≡ φ( ), | | 5 δ−1| | On the other hand, QWQ W ,inranki for Q q2 Q > 6 A (recall that h = δ−1). Therefore H-words V and V are conjugate in rank i by the − product U 1QU. By Lemma 4.11, this product must be equal, in rank i to an H- word X. −1 Let ∆0 be a g-reduced diagram of rank i for the equality (U ) QU = X. Its contour has a natural decomposition (u )−1qux−1. By changing H-word X, one can suppose that there are no T-cells compatible with x in ∆. Hence this section is an ε-section of ∆0 by Lemma 4.14. Therefore the sum of the degrees of contiguity of an arbitrary cell Π of ∆ to q and to x,islessthanα¯ + ε <2/3 by Lemmas 2.3 and Lemma 4.16 below. Recall that Q is a periodic word with the simple in rank i period A. Notice that 5 |u |+|u| <2(βδ−1 + 1)|A| < α2δ−1|A| < α2|q|. 6 NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 101

Thus, there is an 0-bond between q and x by Lemma 2.10 (we have mentioned above that the other possibility from that lemma is not possible). Then q = q(1)eq(2), for the contiguity edge e of the 0-bond. If the length of q(1) or q(2) is greater than 2α−2(βδ−1 + 1)|A|, then one is able to get a new 0-bond between q and x by Lem- ma 2.10. By keeping obtaining new and new such 0-bonds, we conclude, that there is a subpath q(0) of q and a contiguity subdiagram Γ(0) of rank 0 between q(0) and x, where |q(0)| > |q|/2>2|A| > (1 + ε)|A|. Consequently, a reduced form of the H-word X−1 has an A-periodic subword of length > (1 + ε)|A|. By Lemma 4.2(1), the word A is itself freely conjugate to an H-word. This contradicts the deﬁnition of simple words and completes the proof of our lemma.

The proofs of Lemmas 18.7, 18.8 remain unchanged. The formulations of Lemmas 18.9–19.3 [30] are left unchanged. The proofs are also unchanged. Only when we apply our analog of Lemma 18.3 (Lemma 4.7 above) in the proof of Lemma 18.9, we take into account that by T-relations every H-word has order dividing the odd number n in rank i ≥ 1. In the proof of the ﬁrst statement of Lemma 19.3, one should remember that the reduced form of Wn is included in the list of T-relators for an arbitrary H-word W. In order to prove analogs of Lemmas 19.4 and 19.5 [30], we need two more Lemmas 4.13 and 4.14. These two lemmas and our analogs of Lemmas 19.4, 19.5 [30] will be proved by a simultaneous induction on the number of positive cells.

Lemma 4.13. —LetΓ be a contiguity subdiagram of a cell π to a T-cell Π in a g-reduced diagram ∆ of rank i + 1. Then: (1) the rank r(Γ) is 0 and (2) if π = Π, then (π, Γ, Π) < ε.

Proof. — We can assume that the statement of the lemma is not true and the triple (π, Γ, Π) is a minimal counterexample in the sense that |Γ(2)| is minimal possible. Since |Γ(2)| < |∆(2)|, the subdiagram Γ is an A-map by Lemma 4.15. Let p1q1p2q2 = ∂(π, Γ, Π). Notice that |p1|=|p2|=0, because otherwise the principal cell of one of the bonds deﬁning Γ would be a positive cell with degree of contiguity to Π at least ε, which would contradict the minimality of our counterexample. There are no cells in Γ which are compatible with either q1 or q2 because otherwise these cells would be compatible with cells in ∆, and ∆ would be a non-g-reduced diagram. Then q2 is an ε-section by Lemma 4.14(2) for Γ. Also q1 is a smooth section in ∂(Γ) by Lemma 4.16 or 4.14 (2). Therefore the sum of contiguity degrees of any cell of Γ to q1 and q2 is less than α¯ + ε < γ,¯ and so r(Γ) = 0 by Lemma 2.4. 102 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Assume that (π, Γ, Π) ≥ ε. It follows from Lemma 4.2(2) that the cells π an Π are compatible if π is a T-cell. But this is impossible in the g-reduced diagram ∆. If π is not a T-cell, we obtain a contradiction to the deﬁnition of periods by Lemma 4.2(1). Thus the lemma is proved by contradiction.

Lemma 4.14. —LetΓ be a contiguity diagram of a cell π to an H-section q of the boundary of a g-reduced diagram ∆ of rank i + 1. Assume there are no T-cells compatible with q in ∆. Then: (1) r(Γ) = 0 and (2) q is an ε-section of ∂(∆).

Proof. — The proof is analogous to the proof of Lemma 4.13.

The formulation of Lemma 19.4 of [30] is modiﬁed:

Lemma 4.15. — A g-reduced diagram of rank i + 1, having no hoops, is an A-map.

Proof. — The perimeter of any T-cell is at least n by Lemma 4.2 (3), and so condition A1 is true. The proof of part A2 remains the same except at the end the argument related to A-compatibility should be replaced by a similar argument related to the compatibility of a T-cell with an H-section, if the cell Π is a T-cell. The assumption (π, Γ, Π) ≥ ε, in part A3 means that Π is not a T-cell by Lemma 4.13(2), since π = Π (the diagram contains no hoops). Therefore there are no changes in the proof of this part if π is not a T-cell as well. Otherwise the inequality |q2| < (1+γ)|A| follows from Lemma 4.13(1) and Lemma 4.2(1) (because γ > ε).

We add one more possibility to the hypothesis of the analog of Lemma 19.5 in [30]. This lemma is now formulated as follows.

Lemma 4.16. —Letp be a section of the contour of a g-reduced diagram ∆ of rank i + 1 and one of the following possibilities holds: • The label of p is an A-periodic word where A is simple in rank i + 1, • The label of p is an A-periodic word where A is a period of rank k ≤ i + 1,and∆ has no cells of rank kA-compatible with p. • The label of p is an H-word and there are no T-cells in ∆ which are compatible with p. (If p is a cyclic section then in the ﬁrst two cases we further require that φ(p) ≡ Am for some integer m.) Then p is a smooth section of rank |A| in the ﬁrst and the second cases and of rank 1 in the third case, in ∂(∆).

Proof. — We need to prove the two properties from the deﬁnition of a smooth section. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 103

1. The proof that every subpath of p of length ≤ max(k, 2) is geodesic in ∆ coincides with part 1) in the proof of Lemma 19.5 [30] if φ(p) is not an H-word. In the case when φ(p) is an H-word, this property follows from Lemmas 4.13, 2.8 and 4.2. 2. The proof that for every contiguity submap Γ of a cell π to p satisfying (π, Γ, p) ≥ ε we have |Γ p| < (1 + γ)(i + 1) proceeds along the lines of part 2) of Lemma 19.5 [30]. But we need to consider a new case when π is a T-cell. In this case, r(Γ) = 0 as in the proof of Lemma 4.13 (one should just replace Π by p). Then the inequality |q2|≤(1 + γ)(i + 1) follows from Lemma 4.2(1) and from the deﬁnition of T-relations. If φ(p) is an H-word, then the second property of smooth sections follows from Lemma 4.14 (2).

Lemma 4.17. — The constructed group G(∞) satisﬁes identity xn = 1.

Proof. — This immediately follows from Lemma 19.3 (1) [30] (its formulation remains the same in our situation). That lemma states that the order of every word of length at most i,inranki,isadivisorofn.

Proof of Theorem 4.4. — By Lemma 4.5, it remains to show that for arbitrary normal in H subgroup N, the intersection of the normal closure L of N in B(m, n) and H is equal to N, i.e. any H-word w representing an element of L, represents an element of N. Consider the group G(∞) constructed above for this particular N.Sincew represents an element from L in B(m, n), it belongs to the normal closure, in the absolutely free group, of set T plus the set of all words of the form An = 1.ByLem- ma 4.17 and the deﬁnition of G(1) then w represents 1 in G(∞). Therefore there is a g-reduced diagram ∆ of some rank i whose boundary label φ(q) is freely equal to w. By contradiction assume that the H-word w does not represent an element of N, and the diagram ∆ has minimal possible number of positive cells. By Lemma 2.5, there are a cell Π of ∆ and a contiguity subdiagram Γ of rank 0 with (Π, Γ, q) ≥ ε. If Π is a T-cell, then it must be compatible with q by Lem- ma 4.2(2). Therefore when we cut the cell Π from ∆, we multiply (in the free group) the boundary label of ∆ by a word U of the form VWV−1 where W is the boundary label of Π,andV is an H-word. Since W represents an element of N, V represents an element of H,andN is normal in H,thewordU represents an element of N. Hence the boundary label of the resulting diagram ∆1 is an H-word which does not represent an element from N.So∆1 is a counterexample with a smaller number of positive cells. If Π is not a T-cell, i.e. it corresponds to a relation An = 1, then we get a contradiction with Lemma 4.14(2). 104 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

It is clear that if H is a subgroup of B(m, n) satisfying properties (1) and (2) from Theorem 4.4 then any subgroup generated by ﬁnitely many of the free generators of H is also free in the variety of Burnside groups of exponent n, and has the congruence extension property because retracts of subgroups with the congruence extension property also have this property and because the congruence extension property is transitive (if H1 has the congruence extension property in H and H has it in G then H1 has the congruence extension property in G). Theorem 4.4 (and Theorem 1.2) are proved.

4.3. Burnside groups and free products The following lemma is Theorem VI.3.2 from [2] (or Theorem 19.5 from [30]).

Lemma 4.18. — The centralizer of every non-trivial element of B(m, n) is cyclic of order n.

We are going to use the following Lemma repeatedly in Section 6.

Lemma 4.19. —Ifa, b, c ∈ B(m, n), n 1 odd, a = 1, c = 1, [a, c]=1, [bab−1, c]=1 then [b, c]=[b, a]=1.

Proof. — Indeed, suppose that [a, c]=1, [bab−1, c]=1. By Lemma 4.18, the centralizer of every non-trivial element of B(m, n) is cyclic of order n.Hencea, bab−1, c belong to the same cyclic subgroup C of B(m, n). Therefore bCb−1 and C are non- trivially intersecting cyclic subgroups of order n. By Lemma 4.18, bCb−1 = C.Hence b, C is a ﬁnite subgroup of B(m, n) (or order at most n2). By Theorem 19.6 from [30], b, C is cyclic of order n,sob ∈ C.Hence[b, c]=1 as required.

In order to prove Proposition 6.17 below, we will need some properties of free products. Let P and Q be two free groups. Elements of the free product P ∗ Q are represented by words of the form p1q1p2...qsps+1 where pi ∈ P, i = 1, ..., s + 1, is called a P-syllable, qi ∈ Q , i = 1, ..., s, is called a Q -syllable. By deﬁnition, a P-syllable (resp. Q -syllable) is a maximal subword over the set of generators of P (resp. Q ). Elements of P and Q are called 1-syllable elements. Awordp1q1...psqsps+1 from P ∗ Q ,wherepi, qj are syllables, is called *-reduced in P ∗ Q if none of its syllables except possibly for p1, ps+1 is equal to 1. If a product is *-reduced, it is not equal in P ∗ Q to such a product with smaller number of syllables. The P-syllable ps+1 is called the last syllable of p1q1...psqsps+1 (even if ps+1 is empty). If in p1q1...psqsps+1, a syllable, say, pi, is freely equal to 1, we can remove it and glue qi−1 with qi to form a new Q -syllable. Applying this operation several times, we get a *-reduced in P ∗ Q form of the product which is unique. Notice that a word which is *-reduced in P∗Q may not be freely reduced (as a word written in the free generators of P and Q ). NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 105

Let ψ : P → P¯ be a homomorphism of P onto a group P¯ satisfying the identity xn = 1. We say that a word h of P ∗ Q is ψ-trivial if the ψ-image of every syllable of h is 1 in P¯.

Lemma 4.20. —Leth1, h2 be words in P∗Q which are equal modulo Burnside relations. Then the following properties hold.

(i) If h1 is ψ-trivial, and h2 is of minimal length among all words which are equal to h1 modulo Burnside relations then h2 is ψ-trivial.

(ii) Suppose that h1 has exactly two P-syllables p, p with non-trivial ψ-images, h1 ≡ xpyp z, and y is not equal to a word in P modulo Burnside relations. Then h2 is not ψ-trivial. (iii) If h1 is ψ-trivial, h1 = 1 modulo Burnside relations, h2 ∈ P, and modulo Burnside −1 relations h1 = xh2x for some x ∈ P ∗ Q , then there exists a word x1 ∈ P ∗ Q such −1 = −1 ψ that x1h2x1 h1 modulo Burnside relations and the word x1h2x1 is -trivial.

Proof. — (i) Obviously we can assume that h1 is *-reduced. Consider a g-reduced diagram ∆ over the graded presentation of the free Burnside quotient of P ∗ Q from Section 18 of [30] corresponding to the equality h1 = h2. If the rank of ∆ is 0 then h1 = h2 in P∗Q , and each syllable of the word h2 (of minimal length) must be a product of several syllables of h1.Henceh2 is ψ-trivial. ∆ ∂(∆) = φ( ) = φ( −1) = Suppose that the rank of is >0,and s1s2 where s1 h1, s2 h2.Sinces2 is geodesic, s2 is a smooth section of rank |s2|. By Lemma 2.4, part a), and Lemma 2.3 there exists a cell Π of rank >0and its contiguity subdiagram Γ to s1 of contiguity degree >1/2 − γ − α >3ε. By Lemma 2.6, ∆ contains a cell π and a rank 0 contiguity subdiagram Γ of π to s1, such that the contiguity degree of Γ is > ε.Lett be the contiguity arc of Γ on ∂(π), |t| > ε|∂(π)|.Therearetwo possibilities. Case 1. The period of φ(∂(π)) is a 1-syllable word. Then φ(∂(π)) is a 1-syllable word, and one can cut the subdiagram π ∪ Γ from ∆ preserving the ψ-triviality ¯ n of φ(s1).HereweusethefactthatP satisﬁes the identity x = 1. Case 2. The period A of φ(∂(π)) contains at least 2 syllables. Since Γ is a dia- ε φ( )±1 = ±1 gram of rank 0, and n>2, every syllable of A is a syllable of s1 h1 .HenceA is ψ-trivial (since h1 is ψ-trivial). Therefore, again, one can cut the subdiagram Γ ∪ π ±1 1−n from ∆ preserving the ψ-triviality of φ(s1): we replace subword A of φ(s1) by A and reduce the resulting word. Thus in each of these cases we replace ∆ by a diagram of smaller type, and complete the proof by induction on the type of ∆.

−1 −1 (ii) By contradiction assume that h2 is ψ-trivial. We have that pyp = x h2z . −1 −1 By part (i) of this lemma we can replace x h2z by a ψ-trivial word h of minimal 106 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR possible length. Consider a g-reduced diagram ∆ with ∂(∆) = s1t1s2t2 where φ(s1) = p, φ( ) = φ( ) = φ( −1) = ∆ = ∗ t1 y, s2 p , t2 h. If the rank of is 0 then pyp h in P Q .Since y is not equal to a word from P and is ψ-trivial, and ψ(p) = 1,wehavethatthe *-reduced in P ∗ Q form of pyp is not ψ-trivial, a contradiction with the ψ-triviality of h. If the rank of ∆ is >0then as in part (i), we get a cell π and a contiguity subdiagram Γ of π to one of the subpaths s1, t1, s2 with contiguity degree ≥ ε,such that Γ has rank 0. Since φ(s1) and φ(s2) are 1-syllable words, the period A of π is a 1-syllable word if Γ is a contiguity subdiagram to s1 or s2.IfΓ is a contiguity subdiagram to t1, A is a ψ-trivial word. As in part (i) we can remove Γ ∪ π from ∆ preserving the structure of the boundary of ∆ and the condition that φ( y) is a ψ- trivial, and does not belong to P modulo Burnside relations, φ(s1), φ(s2) are 1-syllable words from P, ψ(φ(s1)), ψ(φ(s2)) = 1. We can complete the proof by induction of the type of ∆.

(iii) Consider a g-reduced annular diagram ∆ with pointed contours s1, s2,where −1 −1 φ(s1) ≡ h1,φ(s2) ≡ h2. If the rank of ∆ is 0 then h1 = xh2x in P ∗ Q and xh2x is ψ-trivial. Suppose that the rank of ∆ is >0. As in parts (i), (ii), we can remove a cell of rank >0from ∆ and change s1 or s2 preserving φ(s1), φ(s2) modulo Burnside relations. The labels of the contours s1, s2 of the new annular diagram satisfy all the conditions of part (iii) of the lemma, and again we can complete the proof by induction on the type of ∆. Lemma 4.21. — Suppose that h, g ∈ P ∗ Q ,andthereisnox ∈ P ∗ Q such that g = xpx−1, h = xp x−1 modulo Burnside relations for some p, p ∈ P. Suppose also that h,the *-reduced forms of ghg−1,andawordb which is equal to g−1hg modulo Burnside relations are ψ-trivial. Then g is ψ-trivial.

Proof. — Suppose that g is not ψ-trivial. Let g ≡ g1pg2 where g2 is ψ-trivial, ψ( ) = −1 ≡ −1 −1 −1 p is a P-syllable, p 1. Consider the *-reduced form of ghg g1pg2hg2 p g1 . Since this word is ψ-trivial, and g2 is also ψ-trivial, the syllable p is glued with a P- −1 −1 = −1 syllable ph of h and with the P-syllable p of g . This means that h g2 phg2 in the ∗ −1 −1 −1 free group P Q , and the *-reduced form of ghg is g1pphp g1 .Thisimpliesthatg1 is ψ-trivial, so g has only one P-syllable with non-trivial ψ-image. Now consider the −1 = −1 −1 −1 −1 *-reduced form b of the word g hg g2 p g1 g2 phg2g1pg2. Since b is equal to b modulo Burnside relations and b is ψ-trivial, we can con- −1 −1 ∈ clude by Lemma 4.20(ii) that g1 g2 phg2g1 is equal to a word p P modulo Burn- −1 side relations. Taking the projection of the equality (g2g1) phg2g1 = p onto the sub- −1 group P,wegetP(g2g1) phP(g2g1) = p modulo Burnside relations, where P(g2g1) is −1 the P-projection of g2g1. Therefore P(g2g1) (g2g1) centralizes ph modulo Burnside relations. Since the centralizers of ph in the free Burnside quotient of P and P ∗ Q have NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 107

−1 order n (by Lemma 4.18), these centralizers coincide. Hence P(g2g1) g2g1 belongs to ∈ −1 ∈ P modulo Burnside relations, so g2g1 P modulo Burnside relations. Then g2gg2 P, −1 = −1 −1 = ∈ g2hg2 g2g2 phg2g2 ph P modulo Burnside relations, which is impossible by our assumption.

5. The presentation of the group

In this section, we define an S-machine consisting of a set of (positive) S-rules M+. Using this S-machine, we define our groups H (Theorem 1.3). Fix some numbers first. We use the same odd number n 1 as in the previous sections. Let N>n,andletm = m(n) be the number which exists by Theorem 4.4.

5.1. S-machines and their interpretation

We start with deﬁnition of the generating set X of the group H and the alphabet of the S-machine.

Let A be the set {a1,...,am}.LetΩ be the set {0, ..., 2n + 1}. The set M+ that we shall deﬁne in the next section, will be the union of disjoint + sets Mω , ω ∈ Ω,andtheset{c0, ..., c2n} called the set of connecting rules.Foreach + ω ∈{0, ..., 2n},thesetMω will consist of m rules, one for each letter in A .Therules ∪2n ∪{ , ..., } from ω=0Mω c0 c2n will be called working rules. + + The rules from M2n+1 will be called cleaning rules.ThesetM2n+1 will be in one- to-one correspondence with the set of all working rules (thus the set of cleaning rules and the set of working rules are disjoint but have the same cardinality). Consider the alphabet

−→ ←− K ={λ(i,ω),κ(i), ρ(i), κ (i, j,ω), κ (i, j,ω), i = 1, ..., N, j = 1, ..., n,ω∈ Ω}.

Elements of K will be called K-letters. Let also K be the set of letters obtained from K by removing the Ω-coordinate ω, and let us call the corresponding map from K to K the K-projection. The preimage of every letter z ∈ K under the K-projection will be denoted by K(z).Foreveryz ∈ (K)−1 we denote K(z−1)−1 by K(z). Sometimes it will be convenient to assume that letters with K-projection κ(i) or ρ(i) also have Ω- coordinates but we identify all letters κ(i,ω) with κ(i) and all letters ρ(i,ω) with ρ(i). If U is any word containing letters from K,andω ∈ Ω, then we deﬁne U(ω) as the word, obtained from U by replacing every letter z from K\{κ(i), ρ(i), i = 1, ..., N} by z(ω) (in other words, we add the coordinate ω to each letter from K in U except κ(i), ρ(i)). In this case we shall call ω the Ω-coordinate of U(ω). 108 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

For every z ∈ K\{λ(i), ρ(i), κ( j), 1 ≤ i ≤ N, j = 2, ..., N} consider a copy A(z) of the alphabet A .Foreachz ∈{λ(i), ρ(i), 1 ≤ i ≤ N}, consider the set A(z) which is a copy of the set of working rules: with every working rule τ we associate a letter τ(z) in A(z) (all sets A(z) are disjoint). Notice that we have not defined the sets A(κ( j)), j = 2, ..., N. For convenience we assume that all these sets are empty. Let A be the union of all the sets A(z), z ∈ K. Elements of this set will be called A-letters. Consider also the set of R-letters R ={r(τ, z), τ ∈ M+, z ∈ K}. Now let X = K ∪ A ∪ R. This is the generating set of our group H .WeranksomeofthelettersinX in the following way: K > R > A,thatislettersfromK have rank higher than all other letters, etc. For every word over X we can consider its K-length, etc. Consider the following word Λ over K: −→ ←− −→ ←− λ(1)κ(1)ρ(1) κ (1, 1) κ (1, 1)... κ (1, n) κ (1, n)... (15) −→ ←− −→ ←− λ(N)κ(N)ρ(N) κ (N, 1) κ (N, 1)... κ (N, n) κ (N, n). The word Λ(0) will be called the hub. We define a circular order on K according to the appearance in Λ if we consider the word Λ aswrittenonacircle.Foreveryz ∈ K let z− be the letter immediately preceding z in our order, and let z+ be the letter next to z in that order. − 1 −1 −1 We extend that order also on letters from K .Ifz ∈ K then (z )− ≡ (z+) , −1 −1 −1 (z )+ ≡ (z−) . We define a corresponding circular order on K ∪ K :

z(ω)+ ≡ z+(ω). − 1 −1 For every z ∈ K we set A(z) = A(z− ).Alsobydefinitionforeveryz¯ ∈ K with K-projection z we let A(z¯±1) = A(z±1). The language of admissible words consists of all words of the form y1u1 y2u2...yt where yi ∈ K, ui are words in A( yi), i = 1, 2,...,t − 1,andforeveryi = 1, 2..., ≡ ( ) ≡ −1 either yi+1 yi + or yi+1 yi .ThusallK-letters in an admissible word have the same Ω-coordinates. (Notice that Λ(ω) is an admissible word for every ω.) The subword yiuiyi+1 is called the yi-sector of the admissible word, i = 1, 2..... Two admissible words ...... = ≡ = , ..., y1u1 y2u2 yt and y1u1y2u2 yt are considered equal if t t , yi yi, i 1 t,and = = , ..., − ui ui, i 1 t 1, in the free Burnside group (later on we will identify these ≡ = ∗ admissible words if yi yi and ui ui in the group Hka defined below). The set of defining relations of the group H consists of the hub Λ(0) and relations corresponding to the S-rules. We are going to present these relations now. We start with defining the set of S-rules (an S-machine in the terminology of [36]) and the corresponding defining relations. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 109

Each rule means simultaneous replacing certain subwords in admissible words by other subwords (the rule specifies the subwords which need be replaced and the replacements). We shall write a rule τ in the form [z1 → v((z1)−)z1u(z1), ..., zs → ±1 v((zs)−)zsu(zs); ω → ω ] where zi ∈ K,eachofu(z), v(z) is either a letter from A(z) or empty, z ∈ K. Someofthearrowsinarulecanhavetheform→. This means that the rule can be applied if the corresponding sector does not have A-letters in it. We shall say that the corresponding sectors are locked bytheruleandtherulelocks these sectors. If a z-sector is locked by the rule, u(z) and v(z) must be empty. If z ∈ K does not appear in this rule then we set u(z), v(z−) to be empty. Thus with every rule τ and every z ∈ K we associate two words over A: u(z) and v(z).We − 1 −1 −1 −1 −1 can extend this definition to z ∈ K by saying that v(z ) ≡ u(z) , u(z ) ≡ v(z−) . If for some z ∈ K, u(z) is not empty then we call τ left active for z-sectors, if v(z) is not empty then τ is right active for z-sectors. −1 If τ is a rule as above then for every z ∈ K∪K we define two words L(τ, z) ≡ −1 r(τ, z−)v(z−) and R(τ, z) ≡ r(τ, z)u(z) .Ifh ≡ τ1τ2...τt is a word in the alphabet M, we set L(h, z) ≡ L(τ1, z)L(τ2, z)...L(τt, z), R(h, z) ≡ R(τ1, z)R(τ2, z)...R(τt, z). The rule τ is applicable to a one-letter admissible word y if and only if the Ω- coordinate of y is ω. Let W ≡ y1uy2 be an admissible word with two K-letters. The following two conditions must hold for τ to be applicable to W: (i) y1 and y2 belong to K(ω); (ii) if y1 is one of the letters zi, i = 1, ..., s, τ locks y1-sectors then u must be empty. If conditions (i) and (ii) hold then the result of the application of τ to W is the word ( ) (( ) ) y1u y1 uv y2 − y2 Ω ω ω τ where yi are obtained from yi by replacing the -coordinate by (recall that contains the instruction ω → ω ). If W ≡ y1u1 y2...ut yt is an admissible word with t ≥ 2 K(ω)-letters then the rule τ is applicable to W if and only if it is applicable to every sector of W.Inorderto apply an S-rule to W we apply it to every sector of W. Let W be the resulting word. Notice that W is clearly again an admissible word.

With every rule τ =[z1 → v((z1)−)z1u(z1), ..., zs → v((zs)−)zsu(zs); ω → ω ] we −1 −1 −1 −1 −1 associate the inverse rule τ =[z1 → v((z1)−) z1u(z1) ,...,zs → v((zs)−) zsu(zs) ; ω → ω]. Clearly if τ is applicable to an admissible word W and W is the result of application of τ to W,thenτ −1 is applicable to W and W is the result of application of τ −1 to W . 110 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

The rules from M+, described below, will be called positive, the inverses of these rules will be called negative. Now let us describe the procedure of converting an S-rule into a set of relations. After that we shall present the list of rules M corresponding to our group H .

Let τ =[z1 → v((z1)−)z1u(z1), ..., zs → v((zs)−)zsu(zs); ω → ω ] be an S-rule. Then for every z ∈ K,weintroducearelation

−1 r(τ, z−) z(ω)r(τ, z) = v(z−)z(ω )u(z). If (and only if) τ does not lock z-sectors, we also include all relations of the form r(τ, z)b = br(τ, z), b ∈ A(z).

For every z ∈ K let R(z) denote the set {r(τ, z) | τ ∈ M}. To make formulas look less frightening let us agree on the following:

In every word of the form uar zvar appearing in an S-rule or in a deﬁning relation of our group, or in an admissible word, where z ∈ K(ω), uar , var are words in the alphabet A ∪ R, we shall always assume that the letters in uar are from A(z−) ∪ R(z−) only, and the letters in var are from A(z) ∪ R(z) only. So we shall omit the K-coordinates in these words. We shall also omit the Ω-coordinate in K-letters if the value of it is clear.

Example. — Suppose, for example, that the rule τ has the form −→ −→ ←− ←− [κ(1) → κ(1)a, κ (1, 1) → κ (1, 1)a, κ (1, 2) → a κ (1, 2), 1 → 2] (thisruleisnotinM, we use it just as an example). Then the set of relations corresponding to this rule consists of the following relations. • r(τ)−1κ(1)r(τ) = κ(1)a; − −→ −→ • r(τ) 1 κ (1, 1, 1)r(τ) = κ (1, 1, 2)a; − ←− ←− • r(τ) 1 κ (1, 2, 1)r(τ) = a κ (1, 2, 2); − −→ • r(τ) 1z(1)r(τ) = z(2) for every z ∈ K, z ≡ κ(1), κ (1, 1). ←− • r(τ)b = br(τ) for every b ∈ A(z) and z ≡ κ (1, 2).

− −→ −→ Notice that, for example in the relation r(τ) 1 κ (1, 1, 1)r(τ) = κ (1, 1, 2)a,the two letters r(τ) are different according to our agreement: the ﬁrst letter is r(τ, κ(1)) and the second letter is r(τ, ρ(1)).Thisruleτ is left active for κ(1)-sectors and for −→ ←− ←− κ (1, 1)-sectors, right active for κ (1, 2)-sectors, and locking for κ (1, 1)-sectors. Here is one of the main properties of the systems of relations corresponding to S-rules. The lemma immediately follows from deﬁnitions. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 111

Lemma 5.1. —LetW be an admissible word, starting with z ∈ K and ending with z ∈ K,letτ =[z1 → v((z1)−)z1u(z1), ..., zs → v((zs)−)zsu(zs); ω → ω ] be an S-rule applicable to W (some of the arrows → may be of the form →). Then: 1. L(τ, z)−1WR(τ, z ) is equal modulo the relations corresponding to τ to the admissible word W obtained by applying τ to W. 2. If τ locks z-sectors then the z-sectors of W and W have trivial retraction to the subgroup A in the free group generated by A ∪ K. 3. For every z ∈ K, L(τ, z)z(ω ) = z(ω)R(τ, z) is one of the deﬁning relations corresponding to τ.

5.2. The S-machine M Now let us deﬁne our collection of S-rules M+. As we have announced in the Section 5.1, the set M+ will be divided into 2n + 3 subsets: +,..., + , { ; = , ..., }. M0 M2n+1 ci i 0 2n

The sets Mi will be called steps, ci will be called connecting rules.RulesfromM2n+1 are called cleaning rules, other rules are called working. + ≤ ∈ A In each set Mi , i 2n, each rule corresponds to a letter a ,itwillbe denoted by τ(i, a). The rule τ(0, a) in the set M+ is the following  0  λ( ) → λ( ), κ( ) → κ( ), ρ( ) → ρ( ), −→κ ( , ) → −→κ ( , ) −1,  i i i i i i i j i j a  ←− ←− .  κ (i, j) → κ (i, j),  i = 1, ..., N, j = 1, ..., n; 0 → 0 − −→ This rule inserts copies of the letter a 1 to the right of κ (i, j). A sequence τ( , )±1, ..., τ( , )±1 0 ai1 0 ais ( ±1... ±1)−1 −→κ ( , ) of these rules inserts copies of the word ai1 ais in i j -sectors. These rules lock all other sectors. The connecting rule c has the form 0 ←− ←− λ(i) → λ(i), κ(i) → κ(i), ρ(i) → ρ(i), κ (i, j) → κ (i, j), . i = 1,...,N, j = 1,...,n; 0 → 1 −→ It locks all sectors except for κ (i, j)-sectors. The rule τ(1, a) has the form   −→ −→ κ(i) → τ(1, a)κ(i), κ (i, 1) → τ(1, a) κ (i, 1)a, −→ −→ ←− ←−  κ (i, j) → κ (i, j)a, κ (i, j) → κ (i, j)a, . 1 ≤ i ≤ N, 2 ≤ j ≤ n; 1 → 1 112 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR −→ ←− This rule moves letters from κ (i, j)-sectors to κ (i, j)-sectors (i.e. the rule removes −→ ←− a letter from κ (i, j)-sectors, and adds a copy of that letter to each of the κ (i, j)- sectors). It writes a copy of itself in the λ(i)-sectors and ρ(i)-sectors. It locks all κ(i)- sectors. The connecting rule c1 has the form κ( ) → κ( ), −→κ ( , ) → −→κ ( , ), −→κ ( , ) → −→κ ( , ), i c1 i i 1 c1 i 1 i j i j . 1 ≤ i ≤ N, 2 ≤ j ≤ n; 1 → 2 −→ It writes a copy of itself in the λ(i)-sectors and ρ(i)-sectors. It locks κ (i, j)-sectors and κ(i)-sectors. For every t = 2,...,n and every a ∈ A the rule τ(2t − 1, a) has the form:

  κ(1) → τ(2t − 1, a)κ(1), κ(i ) → τ(2t − 1, a)κ(i ),   −→ −→ ←− ←−   κ (i, 1) → τ(2t − 1, a) κ (i, 1), κ (i, 1) → κ (i, 1),  −→ −→ ←− ←− −→ −→  .  κ (i, s) → κ (i, s), κ (i, s) → κ (i, s), κ (i, j) → κ (i, j)a, ←− ←−  κ (i, j) → κ (i, j)a, 1 ≤ i ≤ N, 2 ≤ i ≤ N, 2 ≤ st,to κ (i, j)-sectors, and writes a copy of itself in the λ(i)-sectors and ρ(i)-sectors. It locks all other sectors. The connecting rule c2t, t = 1, ..., n − 1,hastheform   κ( ) → κ( ), ←κ−( , ) → ←κ−( , )  i c2t i i 1 i 1  −→ −→ ←− ←− .  κ (i, s) → κ (i, s), κ (i, j) → κ (i, j),  1 ≤ i ≤ N, 2 ≤ s ≤ t, 1 ≤ j ≤ n; 2t → 2t + 1

It writes a copy of itself in the λ(i)-sectors and ρ(i)-sectors. It locks all κ(i)-sectors, ←− −→ i>1,all κ (i, j)-sectors and all κ (i, j)-sectors, i ≤ t. The connecting rule c2n, t = 1, ..., n − 1,hastheform   κ( ) → κ( ), −→κ ( , ) → −→κ ( , ), ←κ−( , ) → ←κ−( , )  i c2n i i 1 c2n i 1 i j i j  −→ −→ .  κ (i, j ) → κ (i, j ),  1 ≤ i ≤ N, 1 ≤ j ≤ n, 2 ≤ j ≤ n; 2n → 2n + 1

It writes a copy of itself in the λ(i)-sectors and ρ(i)-sectors. It locks all other sectors. The cleaning rules are in one-to-one correspondence with working rules. For every working rule r, the corresponding cleaning rule τ(2n + 1, r) has the form   κ( ) → κ( ), −→κ ( , ) → −→κ ( , ), ←κ−( , ) → ←κ−( , ),  i r i i 1 r i 1 i 1 i 1  −→ −→ ←− ←− .  κ (i, j) → κ (i, j), κ (i, j) → κ (i, j),  1 ≤ i ≤ N, 2 ≤ j ≤ n; 2n + 1 → 2n + 1

These rules remove the content of λ(i, j)-sectors and ρ(i, j)-sectors (which contain copies of the history of the computation) and locks all other sectors. Notice that only connecting rules change the Ω-coordinates of K-letters. Notice also that for every a ∈ A and every i = 1, ..., 2n, one can deﬁne the rule τ(i, a−1) as before. Similarly for every working rule r one can deﬁne τ(2n + 1, r−1).Itiseasyto see that

τ(i, a−1) = τ(i, a)−1.

The following picture (Figure 13) shows which rules of the S-machine M are locking, active or neither for which sectors of admissible words. A ﬁlled box means that the sector is locked, a box with “>” in it means that the rules are left active for that sector, a box with “<” in it means that the rules are right active, and an empty box means that the rules from the corresponding set of S-rules are neither active for that sector nor locking it. For example, this picture shows that −→ rules from M2 are right active for λ(i)-sectors and ρ(i)-sectors, lock κ (1, 1)-sectors 114 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

FIG. 13.

−→ but are left active for κ (1, 2)-sectors. It also shows that κ(i)-sectors, i ≥ 2,arelocked by all rules of M.

5.3. The presentation The set of relations corresponding to the rules from M will be denoted by Z(M). By Z(M, Λ) we shall denote the set Z(M) together with the hub Λ(0). Our main result is that the group H deﬁned by the set of relations Z(M, Λ) satisﬁes the two conditions of Theorem 1.3. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 115

5.4. The subgroup A of H is a torsion group of exponent n For every group word u in the alphabet A and every K-letter z ∈{λ(i), ρ(i, 1), i = 1, ..., N} we denote a copy of u in the alphabet A(z) by u(z).WemayidentifyA with A(κ(1)).

Lemma 5.2. — For every word u in A ∪ A −1 the relation u(κ(1))n = 1 follows from Z(M, Λ). ≡ ... A Proof. — Let u ai1 ai2 ais be a group word in the alphabet . Start with Λ( )λ( , ) τ( , ) τ( , ) the admissible word 0 1 0 and apply S-rules 0 ai1 ,..., 0 ais from M0 to Λ(0)λ(1, 0). The resulting word is equal to

−→ − ←− −→ − ←− λ(1)κ(1)ρ(1) κ (1, 1)u 1 κ (1, 1) κ (1, 2)u 1 κ (1, 2)... (16) −→ − ←− λ(N)κ(N)ρ(N) κ (N, n)u 1 κ (N, n)λ(1) the Ω-coordinate of each K-letter is 0. Applying now the connecting rule c0, we change the Ω-coordinates of all K- letters to 1. τ( , ) τ( , ) After that, applying rules 1 ais , ..., 1 ai1 ,wegettheword −→ ←− −→ λ(1)v1κ(1)ρ(1)v1 κ (1, 1) κ (1, 1)u κ (1, 2)... (17) −→ −→ ←− λ(N)v1κ(N)ρ(N)v1 κ (N, 1)... κ (N, n) κ (N, n)uλ(1) Ω ≡ τ( , )...τ( , ) (the -coordinates are 1)wherev1 1 ais 1 ai1 . The connecting rule c1 makes all Ω-coordinates 2 and adds letter c1 at the end of v. τ( , ), ..., τ( , ), ,τ( , ), ..., τ( , ), , ..., Now we can apply rules 2 ai1 2 ais c2 3 ais 3 ai1 c3 τ( , ), ..., τ( , ) 2n ai1 2n ais , and obtain the following word:

−n −→ ←− λ(1)v2nκ(1)u ρ(1)v2n κ (1, 1) κ (1, 1)... −→ ←− (18) λ(N)v2nκ(N)ρ(N)v2n κ (N, 1) κ (N, 1)... , −→ ←− κ (N, n) κ (N, n)λ(1) where the Ω-coordinate of every K-letter is 2n, v2n is a copy of the history of the computation. Let W be the word (18) without the last letter. ≡ τ( , )...τ( , ) Let h 0 ai1 2n ais be the string of all rules from M applied so far to Λ(0)λ(1). Then by Lemma 5.1 (part 3), we have the following equality:

L(h, λ(1))Wλ(1, 2n) = Λ(0)λ(1, 0)R(h, λ(1)) = Λ(0)L(h, λ(1))λ(1, 2n). 116 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

W = L(h, λ(1))−1Λ(0)L(h, λ(1)) = 1 modulo Z(M, Λ). Notice that if we apply connecting rules c0, ..., c2n to Λ(0)λ(1, 0),weobtain Λ(2n + 1)λ(1, 2n + 1).

Now take the word Λ(2n + 1)λ(1, 2n + 1) and apply the cleaning rules τ(2n + 1, t1)... τ( + , ) ... ≡ −1 2n 1 ts where t1 ts v2n,andtherulec2n .Wegettheword −→ ←− λ(1)v2nκ(1)ρ(1)v2n κ (1, 1) κ (1, 1)... −→ ←− (19) λ(N)v2nκ(N)ρ(N)v2n κ (N, 1) κ (N, 1)... , −→ ←− κ (N, n) κ (N, n)λ(1) with Ω-coordinate 2n.LetW be the word (19) without the last letter. Let h be the word of rules applied to Λ(0)λ(1, 0) to get (19). Then by Lemma 5.1, we have

L(h , λ(1))W λ(1, 2n) = Λ(0)λ(1, 0)R(h , λ(1)) = Λ(0)L(h , λ(1))λ(1, 2n) modulo Z(M, Λ).ThisimpliesW = 1 modulo Z(M, Λ). But the words W and W differ only by the factor u(κ(1))−n.Thusu(κ(1))n = 1 modulo Z(M, Λ).Hencethe n-th power of every word over A(κ(1)) is 1 modulo Z(M, Λ).

6. Properties of the S-machine M

6.1. The inverse semigroup P(M) We ﬁx an arbitrary set R(κ(1)) of words in the alphabet A(κ(1)) = A . (We choose R(κ(1)) =∅for the construction of the group H , but later we should consider arbitrary R(κ(1)) to prove part (2) of Theorem 1.3.) Let Ha be the group generated by the set A subject all relations of the form n u = 1 (u is any word in A)andrelationsr = 1 for all r ∈ R(κ(1)).ThegroupHa is isomorphic to the free product in the variety of groups of exponent n of subgroups A(z), z ∈ K,whereallA(z), z = κ(1), are free Burnside groups of exponent n, and A(κ(1)) is a group of exponent n with additional relators from R(κ(1)). ∗ Let Hka be the free product of the group Ha and the free group freely generated by the set K. ∗ From now on we will not distinguish admissible words which are equal in Hka. Thus a word of the form y1u1...yt where yi ∈ K, ui are words in A(zi),isadmissible if = , ..., − ≡ ( ) ≡ −1 for every i 1 t 1,eitheryi+1 yi + or yi+1 yi . NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 117

∗ If no sector of an admissible word W is equal in Hka to a word without K-letters then W is called a reduced admissible word.ByLemma3.3thispropertyofanadmissible ∗ word is equivalent to being K-reduced in Hka. The result of an application of an S-rule to an admissible word is again an admissible word. Let W be a word obtained from W by an application of a rule τ from M. Then the K-letters in W differ from the corresponding K-letters in W by their Ω- coordinates only. Therefore we have the following simple fact.

Lemma 6.1. —IfW is obtained from W by a nonempty sequence of applications of S-rules from M then the word W is also admissible and the K-letters of W differ from the corresponding K-letters in W only by the Ω-coordinate. The Ω-coordinate of W is ω if and only if the last rule −1 applied was from Mω or equal to cω−1, cω .

Thus every τ ∈ M deﬁnes a partial one-to-one transformation of the set of admissible words. Let P(M) be the subsemigroup generated by these partial transformations in the inverse semigroup of all partial one-to-one transformations of the set of admissible words. Then P(M) is an inverse semigroup because for every rule τ in M, the inverse rule τ −1 is also in M.ThezeroofP(M) is a transformation with empty domain. Removing zero from P(M) we get what is usually called a pseudogroup of partial transformations. Every word in M induces a partial one-to-one transformation from P(M). The following obvious lemma is a general property of inverse semigroups [8].

−1 Lemma 6.2. —Letawordh over M be graphically equal to h1aa h2. Then, considered as a transformation of the set of admissible words, the domain of h is contained in the domain of h1h2, but the partial transformations induced by h and h1h2 coincide on the domain of h.Thusfor every word h over M the domain of the transformation induced by h is contained in the domain of transformation induced by the reduced form of h.

Notice that the domain of the reduced form of h may be strictly bigger than the domain of h. If h is a word in M then denote by W · h (W is an admissible word), the result of the application of the freely reduced form of h to W (if it exists). Thus the equality W · h = W means that W is in the domain of the freely reduced form of h and the transformation induced by the reduced form of h takes W to W . The following easy but important lemma immediately follows from the deﬁnition of application of an S-rule to an admissible word.

Lemma 6.3. —LetW be an admissible word. Let h ≡ τ1...τt beawordinM.Then the following conditions are equivalent: 118 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR (i) W belongs to the domain of h; (ii) Every sector of W belongs to the domain of h. = · Moreover W W h if and only if for every z-sector W1 of W and z-sector W1 of W = · ∈ ∪ −1 we have W1 W1 h, z K K .

−1 For every τ ∈ M and z ∈ K ∪ K we have defined two words L(τ, z) and R(τ, z). These words belong to the free group generated by A ∪ R.LetLa(τ, z) and Ra(τ, z) be the A-projections of L(τ, z) and R(τ, z). For every word h in the alphabet M,wedefinethewordsL(h, z), R(h, z), La(h, z) and Ra(h, z) in the natural way (L(h, z) and R(h, z) have been already defined before). Clearly La(. , z) and Ra(. , z) are homomorphisms from the free semigroup generated by M into the subgroups A(z−) and A(z) respectively. We say that a word w is a copy of a word w if w is obtained from w by replacing equal letters for equal letters and different letters for different letters. For example, xyx is a copy of aba. The following lemma immediately follows from the definition of the words L(τ,z) and R(τ, z) and the definition of an application of an S-rule.

Lemma 6.4. —(i)Leth be a word over M applicable (in P(M)) to a one-sector admis- · = ( a( , ))−1 a( , ) sible word z1wz2 then z1wz2 h z1 R h z1 wL h z2 z2 where zi differs from zi only by the Ω-coordinate, i = 1, 2. −1 (ii) For every ω ∈ Ω and every z ∈ K ∪ K either La(τ, z) (resp. Ra(τ, z))isemptyfor a a every τ ∈ Mω or L (τ, z) (resp. R (τ, z)) is not empty for every τ ∈ Mω.Ifforsomeτ ∈ M − 1 a a and some z ∈ K ∪ K , L (τ, z) is not empty (resp. R (τ, z+) is not empty) then for every a a τ ∈ M, L (τ, z+) is empty (resp. R (τ, z) is empty). (iii) Suppose that either none of the rules appearing in a word h over M are from M2n+1 or all of them are from M2n+1. Then if all rules of a word h over M are left active for z-sectors for some z ∈ K then Ra(h, z) is a copy of h written in the alphabet A(z) ∪ A(z)−1. If all rules of a awordh are right active for z-sectors for some z ∈ K then L (h, z+) is a copy of h written in the alphabet A(z).

≡ s Lemma 6.5. —(i)Ifh h1 is a reduced word in M, s>1, W belongs to the domain of h then W · h and W have the same Ω-coordinates.

(ii) Let h1 and h2 be two words in M which are equal in the free group modulo Burnside relations. Suppose that W is in the domain of both h1 and h2.ThenW · h1 = W · h2. (iii) Let h beawordinM containing subword g which is equal to 1 modulo Burnside relations: h = h1gh2. Suppose W is in the domain of h.ThenW is in the domain of h1h2 and W · h1h2 = W · h. (iv) If W belongs to the domain of h2,whereh is a cyclically reduced word, then it belongs to the domain of hs for every integer s (in particular, hs has a non-empty domain). NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 119

Proof. — (i) Recall that non-connecting rules do not change Ω-coordinates and each connecting rule cω replaces ω by ω+1, ω ∈ Ω.Henceforeverywordg over M, there exists ω in Ω such that the domain of g consists of admissible words with Ω- coordinate ω.SinceW · h1 is in the domain of h1,theΩ-coordinate of W · h1 is the same as the Ω-coordinate of W.Henceforeveryω ∈ Ω, the product of occurrences ±1 = s ∈ of cω in h1 is1.Thesameistrueforanypowerh h1, s Z,ofh. Therefore for W · h has the same Ω-coordinate as W.

(ii) Let W be in the domain of h1, h2. Suppose that h1 is equal to h2 modulo the Burnside relations in the free group. Clearly it is enough to assume that W has only a −1 a one sector: W ≡ z1wz2. Then by Lemma 6.4(i) W · hi = z1,i(R (hi, z1)) wL (hi, z2)z2,i, a a i = 1, 2.SinceR (. , z1) and L (. , z2) are homomorphisms and h1 and h2 are equal a a a a modulo Burnside relations, R (h1, z1) = R (h2, z1) and L (h1, z2) = L (h2, z2) modulo a −1 a Burnside relations. But all Burnside relations hold in A,so(R (h1, z1)) wL (h1, z2) a −1 a = (R (h2, z1)) wL (h2, z2).

It remains to show that the Ω-coordinates of W · h1 and W · h2 are the same. These Ω-coordinates are determined by the projections of h1 and h2 on the alphabet {c1, ..., c2n+1} (because other rules do not change the Ω-coordinate). Each connecting rule cω induces a partial transformation on Ω of the form ω → ω+1. Therefore every product of cω, ω ∈ Ω which has a non-empty domain considered as a transformation ±1 of Ω, must be freely equal to a word of the form (cici+1...cs) for some i ≤ s.

Since h1 and h2 are equal modulo Burnside relations, their projections are also ±1 equal modulo Burnside relations. But words of the form (cici+1...cs) can be equal modulo Burnside relations only if they are graphically equal. Thus projections of h1 and h2 onto the alphabet {c1, ..., c2n+1} are freely equal. It remains to observe that if two words in cω, ω ∈ Ω, are freely equal and induce non-empty transformations of Ω, then they induce the same transformation of Ω.HenceW · h and W · h have the same Ω-coordinates.

(iii) Indeed by part (ii) W · h1 is equal to W · h1g and is in the domain of h2. Therefore W is in the domain of h1h2 and W · h1h2 = W · h. (iv) Let W belong to the domain of h2 where h is cyclically reduced. It is enough to consider the case when W consists of one sector: W ≡ z1wz2.

Suppose ﬁrst that h does not contain rules which lock z1-sectors. By Lemma 6.4(i) · 2 = ( a( 2, ))−1 a( 2, ) W h z1 R h z1 wL h z2 z2 where z1 and z2 differ from z1 and z2 only by Ω ≡ ≡ · 2 the -coordinate. By part (i), z1 z1 and z2 z2.SoW h is in the domain of h, because W · h2 has the same Ω-coordinate as W,andh does not contain rules which s lock z1-sectors. Thus W is in the domain of h for all positive integers s. By part (ii), W · hn = W. Therefore W is in the domain of h−2 (n>2). This implies as before that W is in the domain of hs for all negative integers s.

Now suppose that h contains a rule τ which locks z1-sectors, h ≡ h1τh2 for some , · , , words h1 h2 in M.ThenW h1 has the form z1z2 where z1 z2 differ from z1 z2 only 120 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

2 by the Ω-coordinate, h ≡ h1τh2h1τh2.ThenW · h1 and W · h1τh2h1 have the same Ω-coordinates because both words are in the domain of τ.Sinceτ locks z1-sectors, · · τ ∗ · W h1 and W h1 h2h1 are both equal to the word of the form z1z2 in Hka.SinceW h1 is in the domain of τh2h1τh2, W·h1τh2h1 must be in the domain of τh2h1τh2 too. Thus 3 2 W is in the domain of h1τh2h1τh2h1τh2 ≡ h . Considering consequently W· h, W· h , ... instead of W, we conclude that W is in the domain of hs for every positive integer s. Using part (ii), as before, we deduce the same fact for negative s.

Lemma 6.6. — For every admissible word U with Ω-coordinate ω, and every two non- empty words h1, h2 over Mω if W is in the domains of h1, h1 = h2 modulo Burnside relations then U · h1 = U · h2. Proof. — By Lemma 6.3, we can assume that U ≡ zuz is a one-sector admissible word. Recall that either all rules from Mω are left (resp. right) active for z-sectors or all these rules lock z-sectors or none of these rules are active or locking for z- sectors. Hence we can conclude that since U is in the domain for h1,andh1 is not empty, it is also in the domain for h2. The equality W · h1 = W · h2 follows from Lemma 6.5 (ii).

Words in the alphabet M, that do not have subwords which are equal to 1 modulo Burnside relations, will be called Burnside-reduced. ≡ ≡ −1 Lemma 6.7. —LetW z1wz2 be an admissible word consisting of one sector, z2 z1 . a a Let h beawordinM. Suppose that R (τ, z1) (resp. L (τ, z2)) is non-empty for all τ in h. Suppose that either none of the rules in h is from M2n+1 or all of them are from M2n+1. Suppose −1 also that W · h = W. Finally suppose that z1 = κ(1), z1 = ρ(1) .Thenh is equal to 1 modulo Burnside relations.

a Proof. — We only consider the case when R (τ, z1) is not empty because the a a L -case is similar. Recall that by Lemma 6.4 (ii) then L (τ, z1) is empty for every τ a −1 from h. By Lemma 6.4 (i), W · h is graphically equal to z1R (h, z1) wz2 and by Lem- a −1 ma 6.4 (iii) R (h, z1) is a copy of the word h written in the alphabet A(z1) ∪ A(z1) . · = a( , ) = ( ) ∗ Since W h W, R h z1 1 in the subgroup A z1 of Hka. This subgroup is a free Burnside group of exponent n (freely generated by A(z1)). Therefore h is equal to 1 modulo Burnside relations.

∗ The presentation of Hka satisﬁes conditions (Z1) if we consider letters from A as 0-letters and let Y = K, so we can apply Lemmas from Section 3.2. In particular, by Lemma 3.3, an admissible word W is K-reduced if and only if it does not have sectors −1 zuz where u = 1 modulo relations of Ha. Lemma 6.8. —LetW be a K-reduced admissible word which is in the domain of some word h over M.ThenW · h is also K-reduced. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 121

Proof. — Indeed, one only needs to check that no sector in W· h is equal to 1 in ∗ τ Hka. Indeed, it is trivial for sectors of the form zwz+ because the application of do ∗ not change the K-projections of K-letters and the free group K is a retract of Hka. If a sector of W has the form zwz−1 then the corresponding sector in W · h has the form zuwu−1z−1 for some u by Lemma 6.4. This sector is equal to 1 only if uwu−1 = 1. = −1 ∗ But then w 1 and the sector zwz is also equal to 1 in Hka. From now on we consider only K-reduced admissible words.

Lemma 6.9. —LetW ≡ zwz−1 be an admissible word consisting of one sector. Let h be a reduced word in M and W is in the domain of h.Thenh does not contain rules which lock z-sectors.

Proof. — Indeed, suppose h ≡ h1τh2 for some τ ∈ M which locks z-sectors. Then by Lemma 6.4 a a −1 −1 −1 W · h1 = z R (h1, z)w(R (h1, z)) (z ) = z (z ) because W · h1 is in the domain of τ.Thisimpliesthatw = 1 and W = 1 which ∗ contradicts the assumption that W is reduced in Hka. Lemma 6.10. —LetW ≡ zwz−1 be an admissible word consisting of one sector. Let h be awordinM and all rules in h are left active for z-sectors. Suppose that W · h = W.Then (i) Ra(h, z) commutes with w (ii) if in addition z ≡ κ(1) and h either does not contain rules from M2n+1 or contains only rules from M2n+1 then h is equal modulo Burnside relations to the power of a certain word rootω(w) depending only on w and the Ω-coordinate of z.MoreoverW · rootω(w) = W.

Proof. — The ﬁrst statement immediately follows from Lemma 6.4(i). To prove (ii), recall that by Lemma 6.4(iii) Ra(h, z) is a copy of h writteninthealphabetA(z)−1 ( ) ∗ and the subgroup A z of Hka is the free Burnside group of exponent n.ByLem- a ma 4.18 then R (h, z) and w belongs to the same maximal cyclic subgroup w0 of A(z) where w0 depends only on w. Therefore h is equal modulo Burnside relations to a power of a copy of w0 writteninthealphabetM. Denote this copy by rootω(w). Then all rules in rootω(w) are left active for z-sectors, and so W is in the domain of a rootω(w).SinceR (rootω(w), z) = w0 commutes with w,wehaveW · rootω(w) = W.

±1 ∓1 Lemma 6.11. —Leth be a reduced word in M containing a subword of the form cω h cω where h is a word in Mω or Mω+1.LetW ≡ z1wz2 be an admissible word consisting of one −1 sector, z1 ≡ κ(1), ρ(1) . Suppose that cω locks z1-sectors, but rules from h are (left or right) active for z1-sectors. Finally suppose that W is in the domain of h.Thenh is equal to 1 modulo Burnside relations. 122 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR ≡ −1 ≡ Proof. — Indeed, since cω locks z1-sectors, by Lemma 6.9 z2 z1 .Thusz2 (z1)+ by the deﬁnition of admissible words. From the deﬁnition of M, it follows that a a then either R (τ, z1) is empty for all τ ∈ M or L (τ, z2) is empty for all τ ∈ M.With- a a out loss of generality assume that L (τ, z2) is empty for all τ from h .ThenR (τ, z1) a are not empty for τ from h and by Lemma 6.4 (iii) R (h , z1) is a copy of h written in −1 a the alphabet A(z1) ∪ A(z1) . By Lemma 6.4 (i), W · h = z1R (h , z1)z2 = z1z2 because −1 a W · h is in the domain of cω which locks z1-sectors. Therefore R (h , z1) = 1,thush is 1 modulo Burnside relations.

The argument in the proof of Lemma 6.11 can be easily generalized to prove the following statement.

Lemma 6.12. —Leth be a reduced word in M of the form h1h2.LetW ≡ z1wz2 be −1 an admissible word consisting of one sector, z2 ≡ (z1)+, z1 ≡ κ(1), ρ(1) . Suppose that W is in the domain of h and W · h1 = W · h1h2 and that rules from h2 are active for z1-sectors. Then h2 = 1 modulo Burnside relations. Lemma 6.13. —LetW ≡ zuz be a one-sector admissible word and h, g be words over M, W · h = W, W · ghg−1 = W. Suppose that at least one of the following three possibilities holds for all rules τ of h, g simultaneously: (i) Ra(τ, z) is not empty; (ii) La(τ, z ) is not empty; (iii) La(τ, z) and Ra(τ, z) are empty.

Suppose also that either all rules from h, g are not from M2n+1 or all of them are from −1 M2n+1. Finally suppose that z ≡ κ(1), z ≡ ρ(1) . Then either g, h commute modulo Burnside relations or W · g = W.

Proof. — Consider two possible cases.

−1 Case 1. W ≡ zuz+.Sincez ≡ κ(1), ρ(1) , by Lemma 6.4(ii) possibilities (i) and (ii) from the formulation of the lemma cannot both hold. By assumption, we have the following possibilities:

a a Case 1.1. For every rule τ from h, g, R (τ, z) is not empty but L (τ, z+) is empty.

a a Case 1.2. For every rule τ from g, h, L (τ, z+) is not empty but R (τ, z) is empty.

a a Case 1.3. For every rule τ from g, hR(τ, z) and L (τ, z+) are empty.

In Cases 1.1, 1.2, by Lemma 6.7, h = 1 modulo Burnside relations, so g, h commute modulo Burnside relations, as required. In Case 1.3 by Lemma 6.4(i) for every τ in g,wehaveW · τ = W,andso W · g = W, as required. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 123

Case 2. W ≡ zuz−1 (again z ≡ κ(1), ρ(1)−1 by assumption).

Case 2.1. Suppose ﬁrst that for every rule from g, h, Ra(τ, z) is not empty.

In this case by Lemma 6.10 h commutes with ghg−1 modulo Burnside relations which implies by Lemma 4.19 that h commutes with g modulo Burnside relations.

Case 2.2. Suppose now that every rule τ of g, h, Ra(τ, z) (and hence La(τ, z−1))isempty.

In this case again by Lemma 6.4(i), for every τ ∈ Mω we have W · τ = W,and so W · g = W, as required.

6.2. Stabilizers in P(M) of subwords of the hub

For every i = 1, ..., N let Λi be the subword of Λ starting with λ(i) and ending with λ(i + 1) (“N + 1”hereis1). Suppose i>1,andthatΛ (0) is in the domain of a word h over M.Let i ←− Ui(ω) ≡ Λi(0) · h, Ui ≡ λ(i)u1κ(i)ρ(i)u2... κ (i, n)u2n+2λ(i + 1) for some words u1, ..., u2n+2 (notice that the word between κ(i) and ρ(i) is empty, i = 2,...,N because every rule in M locks κ(i)-sectors). Let Uj be the admissible word obtained from Ui by replacing all indexes i in K-letters by j: ←− Uj ≡ λ( j)u1κ( j)ρ( j)u2... κ ( j, n)u2n+2λ( j + 1).

It is easy to see by inspection that rules of M act in the same way on Uj for all j>1. Therefore, an induction on |h| gives us

(20) Λj(0) · h = Uj(ω).

LetusalsodeﬁneU1: ←− U1 ≡ λ(1)u1κ(1)Φ(u1)ρ(1)u2... κ (1, n)u2n+2λ(2) where Φ is the homomorphism from A(λ(1)) to A(κ(1)) which takes every letter τ(ω, a) to a(κ(1))−1 if ω is even and greater than 0, and every other letter to 1. No- tice that the map Φ from A(λ(1)) to A(κ(1)) ∪{1} is indeed extendable to a homomorphism from A(λ(1)) to A(κ(1)) because A(λ(1)) is a free Burnside group and A(κ(1)) satisﬁes all Burnside relations.

Remark 6.14. — Notice that U1 would be a copy of Uj obtained by replacing index 1 for j if not for the κ-sector: the κ( j)-sector of an admissible word contains no A-letters, but κ(1)-sectors may contain A-letters. If the A-subword of the κ(1)-sector ±1 of U1 is trivial (say, if U1 is in the domain of c2n ), then U1 is a copy of Uj. 124 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

≡ ... A ∪ A −1 ω Ω ( ) For every word u ai1 ais over and every even from let Tω u be the word τ(ω, )...τ(ω, ). ai1 ais For every odd ω = 2n + 1 from Ω let ( ) ≡ τ(ω, )...τ(ω, ). Tω u ais ai1

−1 Let z, z be two letters from K, z ≡ z+ or z ≡ z then every sector of an admissible word of the form (z(ω)uz (ω))±1 will be called a sector of the type (form) [zz ].

Lemma 6.15. —Letj>1. Suppose Λj(0) is in the domain of a word of the form hc2n ±1 ( + ) ≡ Λ ( ) · where the word h is Burnside-reduced and does not contain c2n .LetUj 2n 1 j 0 hc2n. Then Λ1(0) is in the domain of hc2n and U1(2n + 1) ≡ Λ1(0) · hc2n.

−1 Proof. — Suppose that h contains a subword of the form cih ci where h con- ≡ −1 − tains no connecting rules: h h1cih ci h2 for some even i from 0 to 2n 2.Then ←− consider the sector W of the type [ κ ( j, n)λ( j + 1)] of Λj(0) · h1ci.Sinceci locks ←− ←− κ ( j, n)-sectors, W has the form κ ( j, n,ω)λ( j + 1,ω) (the A-part of the sectors is trivial). Since W · h is in the domain of c−1, W = W · h . Since all rules in h ←− i are left active for κ ( j, n)-sectors (all these rules belong to Mi+1,andi is odd), by Lemma 6.11, h = 1 modulo Burnside relations, a contradiction. Similarly, h does not −1 contain subwords of the form ci h ci with even i>0where h does not contain connecting rules. −1 Suppose that h contains a subword of the form cih ci where h contains no connecting rules and i is odd, i = 2n − 1. Then applying the argument similar to the −→ ←− one used in the previous paragraph to the sector of the form [ κ ( j, n) κ ( j, n)],we get a contradiction. Similarly one can prove that h does not contain subwords of the −1 form ci h ci where h contains no connecting rules, i is any odd number between 1 and 2n − 1. Suppose now that h contains one of the connecting rules twice. Since Λj(0) has Ω-coordinate 0,andΛj(0) · h has Ω-coordinate 2n, the facts proved in the previous two paragraphs imply that h contains a subword of the form ... −1 ... −1 h0c0h1 h2n−2c2n−1h2n−1c2n−1h2n−2 c0 h0 where the subwords hω, hω are words over Mω. Let then W be the sector of the form −→ [ρ( j) κ ( j, 1)] of Λj (0) · h0.ThewordW is in the domain of ≡ ... −1 ... −1, h c0h1 h2n−2c2n−1h2n−1c2n−1h2n−2 c0 c0 locks W , all other rules from h are right active for ρ( j)-sectors. None of these rules belong to M2n+1. Hence by Lemma 6.11, h = 1 modulo Burnside relations, a contradiction. Therefore h contains each of its connecting rules only once. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 125

Hence h can be represented in the form h0c0h1c1...h2n for some words hi over Mi, i = 0, ..., 2n. −1 For some words u0, ..., u2n over A ∪ A , hi ≡ Ti(ui), i = 0, ..., 2n. Let us prove by induction on i that ui = u0 modulo Burnside relations. Indeed, this statement is trivial for i = 0. Suppose that it is true for some ω from 0 to 2n − 1. Let us prove it for ω + 1.Sinceu0 = u1 = ... = uω modulo Burnside relations, by Lemma 6.6,

Λj(0) · h0c0h1c1...hωcω = Λj(0) · T0(u)c0T1(u)c1...Tω(u)cω = V. In particular the Ω-coordinate of V is ω + 1. Suppose that ω is even. Then as in the proof of Lemma 5.2 the sector of the −→ ←− −→ ←− form [ κ ( j, n) κ ( j, n)] of V is equal to κ ( j, n,ω + 1)u κ ( j, n,ω + 1).SinceV · −→ Tω+ (uω+ ) is in the domain of cω+ and cω+ locks κ ( j, n)-sectors, we get that the 1 1 −→ ←− 1 1 sector of the form [ κ ( j, n) κ ( j, n)] of V · Tω+1(uω+1) is equal to −→ ←− κ ( j, n,ω+ 1) κ ( j, n,ω+ 1).

By Lemma 6.4, we get that uω+1 = u modulo Burnside relations, as required. If ω is odd then we can apply the above argument to the sector of the form ←− [ κ ( j, n)λ( j + 1)] of V. Therefore we can apply Lemma 6.6 2n + 1 times and conclude that Λ1(0) · h0 = Λ1(0) · T0(u), Λ1(0) · h0 is in the domain of c0h1 and Λ1(0) · h0c0h1 = Λ1(0) · T0(u)c0T1(u),..., Λ1(0) · h1c0...h2n−1 = Λ1(0) · T0(u)c0...T2n−1(u) is in the domain of c2n−1h2n and

Λ1(0) · h1c0...h2n−1c2n−1h2n = Λ1(0) · T0(u)c0...T2n−1(u)c2n−1T2n(u) = V . All sectors of the latter word contain no A-letters except for the λ(1)-sector, κ(1)- sector, and ρ(1)-sector. The ﬁrst and the third of these sectors contain copies hc2n n of the word hc2n,theκ(1)-sector contains u (as in the proof of Lemma 5.2) which n is equal to 1 modulo relations of A(κ(1)). Clearly Φ(hc2n) = u .ThusΛ1(0) · hc2n exists and is equal to U1(2n + 1).

Proposition 6.16. — Suppose that for some Burnside-reduced word h over M,andsome j>1, Λj(0) · h = Λj(0).Thenforeveryi = 1,...,N, Λi(0) · h = Λi(0).

Proof. — If h is empty, the statement is obviously true. By (20), the statement is also true if i>1.Soleth be non-empty and i = 1. Then as in the proof of Lemma 6.15, Lemma 6.11 implies that h contains no ±1 ∓1 = , subwords of the form ci f ci where f does not contain connecting rules, i 0 2n. Hence h can be represented in the form −1 −1 −1 −1 −1 −1 ... −1 −1 h1c2ng1c2n h2 c0 h2c2ng2c2n h3 c0 h3c2ng3 gs−1c2n hs = , ..., − , ..., , ,..., where gp are non-empty words over M2n+1, p 1 s 1, h1 hs h2 hs−1. 126 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

We claim that for every t = 2, ..., s − 1,theword

≡ Λ ( ) · −1 −1 −1 −1... −1 −1 −1 V j 0 h1c2ng1c2n h2 c0 h2c2ng2c2n gt−1c2n ht c0

−1 is equal to Λj(0) · T0(ut) for some word ut over A ∪ A . Indeed, let

≡ Λ ( ) · −1 −1 −1 −1... −1. V j 0 h1c2ng1c2n h2 c0 h2c2ng2c2n gt−1c2n

≡ · −1 −1 Then V V ht c0 . By Lemma 6.11, the word ht is represented in the form ... ≡ ( ) = , ..., · −1 p1c1p2c2 p2n where pi Ti vi for some words vi over Mi, i 1 2n.SinceV ht −1 · −1 is in the domain of c0 , V ht has the form −→ ←− −→ λ( j, 1)κ( j, 1)ρ( j, 1) κ ( j, 1, 1)w1 κ ( j, 1, 1) κ ( j, 2, 1)w2... −→ ←− κ ( j, n, 1)wn κ ( j, n, 1)λ( j + 1, 1).

· −1 ( ) −→κ ( , ) Since V ht T1 v1 is in the domain of c1,andc1 locks any j i -sector, , ..., · −1 −1 = by Lemma 6.4(iii), each of the words w1 wn is a copy of v1.HenceV ht c0 Λj(0) · T0(v1) as required. Λ ( ) ( ) ( ) Therefore j 0 is in the domain of T0 ut c0htc2n and T0 ut htc2n. ( + ) ≡ Λ ( )· ( ) ( + ) ≡ Λ ( )· ( ) Let Uj,t 2n 1 j 0 T0 ut c0htc2n, Uj,t 2n 1 j 0 T0 ut htc2n.ByLem- Λ ( ) ( ) ( ) ( + ) = ma 6.15, 1 0 is in the domain of T0 ut c0htc2n and T0 ut htc2n,andU1,t 2n 1 Λ ( ) · ( ) ( + ) = Λ ( ) · ( ) j 0 T0 ut c0htc2n, U1,t 2n 1 j 0 T0 ut htc2n. ( + ) ( + ) ( + ) By Remark 6.14, U1,t 2n 1 is a copy of Uj,t 2n 1 and U1,t 2n 1 is a copy ( + ) of Uj,t 2n 1 An easy induction on the length of a word g over M2n+1 gives that ( + ) · ( + ) · ( + ) ( + ) U1,t 2n 1 g (resp. U1,t 2n 1 g)isacopyofUj,t 2n 1 (resp. Uj,t 2n 1 ). Thus we can conclude that ((...(((Λ ( ) · ) · ( −1 −1 −1 ( )−1)) · ( ) ) · ...)· 1 0 h1c2ng1 c2n h2 c0 T0 u1 T0 u1 h2c2ng2 ( ) ) · −1 −1 = Λ ( ). T0 us−1 hs−1c2ngs−1 c2n hs 1 0

By Lemma 6.2, we obtain Λ1(0) · h = Λ1(0) as required.

6.3. Stabilizers of arbitrary admissible words We call an admissible word W accepted if for some word h over M,thewordW·h is a cyclic shift of Λ(0)±1. Clearly Λ(0) is an accepted word and every admissible word of the form Λ(0) · h is accepted. In the rest of the section, we are going to prove the following statement.

Proposition 6.17. —LetW be an admissible word of the form z1w1z2...zswsz1 (the ﬁrst and the last K-letters are the same) which does not contain an accepted subword. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 127 (1) Suppose that for some i = 1, ..., N, every sector of W is of one of the types [κ(i)−1κ(i)], [κ(i)ρ(i)] or [ρ(i)ρ(i)−1].Leth, g, b be words in the alphabet M such that W · h = W, W · ghg−1 = W, W · b = W and b is equal to g−1hg modulo Burnside relations. s −s Then for every integer s there exists a word bs which is equal to g hg modulo Burnside relations and such that W · bs = W. (2) Suppose that there is no i from 1 to N such that every sector of W is of one of the types [κ(i)−1κ(i)], [κ(i)ρ(i)] or [ρ(i)ρ(i)−1].Leth, g be words in the alphabet M such that W · h = W, W · ghg−1 = W.ThenW · g = W or g and h commute modulo Burnside relations.

Remark 6.18. — Clearly if W · g = W or g and h commute modulo Burnside relations then the conclusion of (1) holds, so the conclusion of (2) is stronger than the conclusion of (1).

Let us prove condition (1) of the proposition. Thus consider the case when for some i = 1, ..., N, W contains sectors only of the form [κ(i)−1κ(i)], [κ(i)ρ(i)] or [ρ(i)ρ(i)−1]. Notice that since W starts and ends with the same K-letter, the deﬁnition of admissible words forces W to have all three of these types of sectors. Indeed, suppose that W contains a sector of the type [κ(i)ρ(i)].ThenW must contain sectors of the types [zκ(i)] and [ρ(i)z ] for some z, z . By assumption, each of these sectors can be of one of the three types [κ(i)−1κ(i)], [κ(i)ρ(i)] or [ρ(i)ρ(i)−1].Hencez ≡ κ(i)−1, z ≡ ρ(i)−1. Thus sectors of all three types are inside W.IfW contains a sector of the type [κ(i)−1κ(i)] then it must contain a sector of the type [κ(i)z] for some z and by assumption z cannot be anything but ρ(i).SoW contains a sector of the type [κ(i)ρ(i)] which implies that W contains all three types of sectors. Finally if W contains a sector of the type [ρ(i)ρ(i)−1] then W contains a sector of the type [zρ(i)] for some z.This z must be equal to κ(i),soW contains a sector of the type [κ(i)ρ(i)] which, as before, implies that W contains all three types of sectors. ∪{ , −1} Therefore by Lemma 6.9 h and g do not contain rules from M0 c0 c0 . We will use this below several times. We are going to use lemmas from Section 4.3. Consider the free group freely \( ∪{ , −1}) generated by M M0 c0 c0 as the free product of the group Q freely generated by M2n+1 ∪{c2n} and the group P freely generated by other generators. Consider the homomorphism υ from P to A(κ(1)) which takes every rule τ from P to Ra(τ, κ(1)). This allows us to talk about syllables of words in P ∗ Q and about υ-trivial elements of P ∗ Q . Notice that if h and the reduced form of ghg−1 consist of rules from P,wecan apply Lemma 6.13 to a sector of the form [κ(i)−1κ(i)] (case (i) of the formulation of that lemma holds) and conclude that g, h commute modulo Burnside relations. This implies part (1) of the proposition. 128 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Thus we can assume that h or the reduced form of ghg−1 contain rules from ∪{ , −1} · = , · −1 = −1 M2n+1 c2n c2n .SinceW h W W ghg W, g or the reduced form of ghg ∪{ , −1} must contain a rule from M2n+1 c2n c2n . Therefore there exists a word x in M such that W = W · x is in the domain of ∪{ , −1} · −1 = ·( −1 )( −1 )( −1 ) arulefromM2n+1 c2n c2n . By Lemma 6.2 W x hx W , W x gx x hx x gx = · −1 = W , W x bx W . It is clear that if for every integer s there exists bs which is equal ( −1 )s( −1 )( −1 )−s · = to x gx x hx x gx modulo Burnside relations and such that, W bs W ,then condition (1) holds for W, g, h. Hence we can assume that W = W . Thus all sectors of the form [κ(i)ρ(i)] in W are equal to κ(i)ρ(i) (since every ∪{ , −1} κ( ) rule from M2n+1 c2n c2n locks i -sectors). a For every i = 1, ..., N, consider the homomorphism ϑi : τ → L (τ, κ(i)) from P ∗ Q to A(λ(i)).

Lemma 6.19. —Letx ∈ P ∗ Q . Then the following conditions hold: (i) κ(1)ρ(1) · x = κ(1)ρ(1) if and only if x is υ-trivial. (ii) κ(1)ρ(1) is in the domain of x if and only if every P-syllable of x except possibly the last one has trivial υ-image. (iii) Let W ≡ κ(i)−1uκ(i) be a 1-sector admissible word, i = 1, ..., N.ThenW · x

= W if and only if ϑi(x) commutes with u modulo Burnside relations. (iv) Every 1-sector admissible word of the form [κ(i)−1κ(i)] or [ρ(i)ρ(i)−1], i = 1, ..., N, is in the domain of x. (v) For every sector W of the form [ρ(i)ρ(i)−1] of the word W, W · x = W . (vi) If i = 1 then the word W is in the domain of x if and only if all P-syllables of x except possibly the last one are υ-trivial. If i>1then the word W is in the domain of x. (vii) If i = 1 then W · x = W if and only if x is υ-trivial, and for every sector W ≡ −1 κ(1) uκ(1) of W, ϑ1(x) commutes with u modulo Burnside relations. If i>1then −1 W · x = W if and only if for every sector W ≡ κ(i) uκ(i), ϑi(x) commutes with u modulo Burnside relations.

Proof. — Sinceweidentifylettersκ(i,ω), ω ∈ Ω,withκ(i) and ρ(i,ω), ω ∈ Ω, with ρ(i) (i = 1, ..., N), all rules from P are applicable to W and to any word of the form W · x.RulesfromQ are applicable to such a word if and only if every sector [κ( )ρ( )] κ( )ρ( ) ∗ of the form 1 1 is equal to 1 1 in Hka. This implies parts (i), (ii) of the lemma. All rules from P ∗ Q are applicable to any sector of the form [κ(i)−1κ(i)] or −1 a a [ρ(i)ρ(i) ].Foreveryruleτ from P ∗ Q , R (τ, ρ(i)) = L (τ, ρ(i)+) = 1.Thisimplies parts (iii), (iv) and (v) because of Lemma 6.4. Parts (vi), (vii) follow from the previous parts of the lemma by Lemma 6.3. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 129

Lemma 6.20. —LetU ≡ κ(i)−1uκ(i) be a one-sector admissible word i ≥ 1. Suppose that U · h = U, U · ghg−1 = U. Then for every positive integer s, U · gshg−s = U.

−1 Proof. — By Lemma 6.19 (iii) ϑi(h), ϑi(ghg ) commutes with u modulo Burnside relations. Since u = 1 modulo Burnside relations (we have agreed to consider only reduced admissible words), by Lemma 4.19, then ϑi(g) commutes with ϑi(h) modulo Burnside relations. Hence ϑ(gshg−s) = ϑ(h) modulo Burnside relations. Since by Lemma 6.19 (iv) U is in the domain of gshg−s, by Lemma 6.5(ii), we get U · gshg−s = U · h = U.

Now we are ready to prove part (1) of the proposition if i>1.

Lemma 6.21. — Part (1) of the Proposition 6.17 holds if i>1.

Proof. — Indeed all rules from M lock sectors of the form [κ(i)ρ(i)]andarenot left active for ρ(i)-sectors and for ρ(i)−1-sectors. Therefore if W is any of the sectors of W of the form [κ(i)ρ(i)]or[ρ(i)ρ(i)−1]thenW · g = W , so for every integer s, W · gshg−s = W . By Lemma 6.20 the same conclusion can be drawn for sectors of the form [κ(i)−1κ(i)]. Hence by Lemma 6.3 W · gshg−s = W as required.

Now let i = 1. It is obvious that we can assume that g, h = 1 modulo Burnside relations.

Lemma 6.22. — If there exists a word x ∈ P ∗ Q such that g = xpx−1, h = xqx−1 modulo Burnside relations, where p, q ∈ P,theng and h commute modulo Burnside relations.

Proof. — Suppose that such a word x exists. By Lemma 6.19 (i), the word h, the reduced form of ghg−1 and the word b from the formulation of part (1) of the proposition are υ-trivial. By Lemma 4.21, g is also υ-trivial. Then by Lemma 4.20(iii) , = −1 = −1 = (since g h 1)thereexistwordsx1 and x2 such that x1px1 g, x2qx2 h mod- −1 −1 υ , = ulo Burnside relations, x1px1 , x2qx2 are -trivial. Since h g 1 modulo Burnside −1 −1 υ( ) = relations, x1px1 , x2qx2 can be considered reduced. This implies that p υ(q) = 1. −1 −1 We have that x x1 commutes with p modulo Burnside relations. Therefore x x1 belongs to the centralizer of p modulo Burnside relations. The centralizer of every non-trivial element of a free Burnside group is cyclic of order n by Lemma 4.18. Therefore the centralizer of p in the free Burnside factor of P ∗ Q and in the free −1 Burnside factor of P are the same. Thus x x1 belongs to P modulo Burnside rela- = ∈ −1 tions and x x1p1 for some p1 P from the same cyclic subgroup as p.Sincex1px1 is ∗-reduced and υ-trivial, all P-syllables of x1p1 except for, possibly, the last one are υ-trivial. Hence by Lemma 6.19(vi), W is in the domain of x1p1. 130 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

−1 −1 = −1 = Notice that x1p1qp1 x1 xqx h modulo Burnside relations. Hence by Lem- · −1 −1 = · = −1 ma 6.5(ii) W x1p1qp1 x1 W h W.Thewordghg is equal to

−1 −1 −1 −1 −1 = −1 −1 −1 x1px1 x1p1qp1 x1 x1p x1 x1pp1qp1 p x1 modulo Burnside relations. = · · −1 = · ( −1) −1 = Let W1 W x1.ThenW1 p1qp1 W1 and W1 p p1qp1 p W1.By [κ( )−1κ( )] −1 Lemma 6.10 applied to a i i -sector of W1, p commutes with p1qp1 modulo Burnside relations. Conjugating by x1, we deduce that g, h commute modulo Burnside relations.

Now we are ready to prove part (1) of Proposition 6.17 for i = 1. Recall that we can assume that all sectors of W of the form [κ(1)ρ(1)] are equal κ( )ρ( ) ∗ to 1 1 in Hka. By Lemma 6.22, we can assume that there is no word x such that g = xpx−1, h = xqx−1 modulo Burnside relations, where p, q ∈ P. By Lemma 6.19(i), h, ghg−1,andb are υ-trivial words. By Lemma 4.21 then g is an υ-trivial word. Hence by Lemma 6.19(i), W1 · g = W1 for every sector of W of s −s the form [κ(1)ρ(1)].ThisimpliesthatW1 · g hg = W1 for every integer s.ByLem- ma 6.20, the same equality is true for every sector of W of the form [κ(1)−1κ(1)].By Lemma 6.19(v) the same is true for sectors of the form [ρ(1)ρ(1)−1].ThusW · gshg−s = W for every integer s. This completes the proof of part (1) of Proposition 6.17. Now let us prove part (2) of the proposition. So let us assume that there is no i = 1, ..., N such that each sector of W is of one of the forms [κ(i)ρ(i)], [κ(i)−1κ(i)], [ρ(i)ρ(i)−1]. Assume that part (2) of Proposition 6.17 is not true and that (|h|, |g|) is the smallest in lexicographic order pair of numbers for all such counterexamples (h, g) (for any W). Clearly the word h is not empty. Thenextfourlemmasallowustoimproveoutcounterexample(h, g).

Lemma 6.23. — For every word x over M such that W is in the domain of x, the pair (x−1hx, x−1gx) is also a counterexample to statement (2) of Proposition 6.17.

Proof. — Indeed by Lemma 6.2 (W · x) · x−1hx = W · x,

(W · x) · (x−1gx)(x−1hx)(x−1gx)−1 = W · x.

Also since W does not contain accepted subwords, W · x does not contain accepted subwords.

Lemma 6.24. —Thewordh is cyclically reduced. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 131

Proof. — Suppose that h is not cyclically reduced. We can represent h in the form h1h2 such that the reduced form h of the word h2h1 is cyclically reduced and | | | | ( , −1 ) h < h . By Lemma 6.23, the pair h h1 gh1 is also a counterexample to part (2) of Proposition 6.17. This contradicts the minimality of the counterexample (h, g).Hence h is cyclically reduced.

Lemma 6.25. — There exists a possibly different counterexample (h¯, g¯) with |h¯|=|h|, |g¯|=|g| such that h¯ is cyclically reduced and g¯h¯g¯−1 is reduced. Moreover for some preﬁx x−1 of the word h, we have h¯ = xhx−1, g¯ = xgx−1 in the free group.

Proof. — By Lemma 6.24, h is cyclically reduced. Therefore either gh or hg−1 is a freely reduced word. Let us assume that hg−1 is freely reduced. The right-left dual of the argument presented below will work in the case when gh is freely reduced. ≡ −1 −1 −1 Notice that g g1h for any g1, since otherwise g1hg1 is freely equal to ghg , and the counterexample (h, g1) would be smaller than (g, h).Thusg ≡ g1 f where −1 ≡ −1 = −1 −1 f f h and g1 f is a reduced word. Hence ghg g1 f f g1 , whence · −1 −1 = . (21) W g1 f f g1 W Since W · f −1f = W · h = W,wehave(W · f −1) · f f −1 = W · f −1.LetW = W · f −1. −1 −1 −1 Then W · f f = W and W · ( fg1)f f ( fg1) = W by (21). −1 Now notice that since h is cyclically reduced, | f f |=|h|.If| fg1| < |g| then −1 the minimality of (|h|, |g|) implies that W · ( fg1) = W or fg1 commutes with f f modulo Burnside relations. In the ﬁrst case W · g1 f = W.SinceW · h = W,weget −1 W · g = W · (g1 f ) = W, a contradiction. In the second case conjugating by f ,we −1 get that g1 f commutes with f f = h modulo Burnside relations. Therefore g ≡ g1 f commutes with h modulo Burnside relations, a contradiction. Hence the length of the freely reduced form of fg1 is equal to |g|.Thisimplies that if g1 is not empty then g is cyclically reduced (since f is not empty by our assumption).

−1 −1 −1 Case 1. Suppose that g1 is not empty. Then the word fg1 f f g1 f is reduced −1 ( fg1 is reduced because g ≡ g1 f , | fg1|=|g|, g1 f is reduced by the deﬁnition, f f is −1 −1 reduced because it is a cyclic shift of the cyclically reduced word h, f g1 is reduced ¯ −1 because g1 f ≡ g is reduced). In this case h ≡ f f , g¯ ≡ fg1 and x ≡ f . Therefore W is in the domain of x−1. By Lemma 6.23, the pair (h¯, g¯) is a (minimal) counterexample, as required.

Case 2. Now let g1 be empty, that is f ≡ g.Ifg is not cyclically reduced, then ≡ −1 ≡ −1 −1 , g yg1 y , h yg2 y h2 for some g2 h2 and a non-empty y. Applying Lemma 6.23 −1 for x ≡ y, we obtain a counterexample (g1 y h1 y, g2).Sinceg2 is shorter than g,and 132 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

−1 g1 y h1 y isofthesamelengthash, we get a contradiction. Hence g is cyclically reduced.

Now let g−s be the maximal power of g which is a prefix of h.UsingLem- ma 6.23 with x ≡ g−s, we can pass from the pair (h, g) to another minimal counterex- ˆ −s −s ample (h, g) = (h1g , g) where h1 is defined by the formula h ≡ g h1.Noticethat −s since h is cyclically reduced by Lemma 6.24, h1g is cyclically reduced. The word h1 is not trivial because otherwise g and h would commute, so (h, g) would not be a counterexample. Since g is cyclically reduced, the product hgˆ −1 is reduced. If ghˆ is reduced, the product ghgˆ −1 is reduced and we are done. Otherwise, since g−1 is not a prefix of hˆ (by the choice of s), we can apply the argument of Case 1 and produce the required counterexample.

Lemma 6.25 allows us to assume (from now on) that ghg−1 is reduced.

Lemma 6.26. —Thewordsg, h are Burnside reduced.

Proof. — Let h contain a non-empty subword c which is equal to 1 modulo Burn- side relations. Then h ≡ h1ch2 for some h1, h2. By Lemma 6.5 (iii), W·h1h2 = W.Since −1 −1 −1 ghg is reduced, we have W · g · h1ch2 · g = W,soW · gh1h2g = W. Now it is clear that (h1h2, g) is a counterexample to part (2) of Proposition 6.17, which contradicts the minimality of (h, g). Similarly if g is not Burnside-reduced, we can shorten g.

= −1 Lemma 6.27. — One cannot represent h as a product h1h2h3 where h1 h3 modulo Burnside relations and h1, h3 consist of rules from the same Mω, ω ∈ Ω, and at least one of the words h1 or h3 is not empty.

Proof. — If one of the words h1 or h3 is empty then h1 = h3 = 1 modulo Burnside relations, and we get a contradiction with Lemma 6.26. Hence both h1 and h3 are not , −1 , −1, empty. Let f be a shortest word among h1 h3 . By Lemma 6.6, h1 h3 f deﬁne the same transformation from P(M).Hencewehave

−1 −1 W · h = W · h1 · h2 · h3 = W · f · h2 · f = W · fh2 f = W and, since ghg−1 is reduced,

−1 −1 −1 W · g · h1 · h2 · h3 · g = W · g( fh2 f )g = W.

−1 −1 Since h = fh2 f modulo Burnside relations, the pair ( fh2 f , g) is a counterexample −1 to part (2) of Proposition 6.17. Since | f |≤|h1|, |h3|, ( fh2 f , g) is a minimal coun- −1 terexample. But fh2 f is not cyclically reduced, a contradiction with Lemma 6.24. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 133

−1 ±1 Lemma 6.28. — Suppose that ghg contains a connecting rule cω which locks a sector −1 −1 ±1 zwz of W, z ≡ κ(1), ρ(1) . Suppose that all rules in ghg which differ from cω are left or −1 ±1 ±1 right active for z-sectors. Then ghg contains at least two different connecting rules ci and ci+1, i ∈ Ω.

−1 Proof. — By Lemma 6.9, z ≡ z+; z ≡ κ(i), ρ(i) for any i because all rules lock κ(i)-sectors if i>1,andz ≡ κ(1), ρ(1)−1 by assumption. Hence every connecting rule changes the Ω-coordinate of z-sectors. Suppose, by contradiction, that ghg−1 does not contain two different connecting rules. Since W · ghg−1 = W and W · h = W,andthewordghg−1 is reduced, W and ±1 W · g have the same Ω-coordinate. Hence either g does not contain cω or g contains ±1 ∓1 a subword cω f cω where f contains no connecting rules. In the first case, h contains ±1 ±1 ∓1 cω ,so(sinceW · h = W), h contains a subword of the form cω f cω .Inbothcases f contains rules from only one Mi. All these rules are active for z-sectors by our assumption. Hence by Lemma 6.11 (which can be applied since z ≡ κ(1), ρ(1)−1), f = 1 modulo Burnside relations, a contradiction with Lemma 6.26. −→ −→ − Lemma 6.29. — W cannot contain a sector of the form [ κ (i, j) κ (i, j) 1]. −→ −→ − Proof. — Suppose that W contains such a sector W ≡ κ (i, j,ω)w κ (i, j,ω) 1. Notice that since W starts and ends with the same K-letter, it must contain a sector W of the form z−1w z, z ∈ K(ω), either to the left or to the right of our sector W . Replacing W by W−1 if necessary, we can assume that such a sector occurs to the left of W . We also choose W to be as close to W on the left as possible. −1 By Lemma 6.9, ghg does not contain occurrences of cω or rules from Mω −→ −→ which lock κ (i, j)-sectors. For every t ∈ Ω either ct or ct+1 locks κ (i, j)-sectors, −1 and c2n and every rule from M2n+1 lock these sectors. Therefore ghg can only contain rules from two consecutive Mt, Mt+1 (ω = t or ω = t + 1), and does not contain ∪{ , −1} rules from M2n+1 c2n c2n . −1 If ghg contains rules only from one Mt, then by Lemma 6.10(ii), we have that h −1 and ghg are equal modulo Burnside relations to powers of rootω(w). By Lemma 4.19, it follows that h and g commute modulo Burnside relations, a contradiction. −1 Thus we can assume that ghg contains rules from two different sets Mt and −1 ≡ ±1 ∓1 −1 Mt+1 or ghg ct f ct for some f .Inbothcasesghg contains an occurrence of −1 ct and an occurrence of ct . Since W is not a prefix of W, by the definition of admissible words, W contains a subword of one of the following forms: −→ −1 −→ −→ −1 κ (i, j,ω) w1 κ (i, j,ω)w κ (i, j,ω) , ←− −→ −→ −1 κ (i, j − 1,ω)w1 κ (i, j,ω)w κ (i, j,ω) ( j>1), −→ −→ −1 ρ(i)w1 κ (i, 1,ω)w κ (i, 1,ω) , −→ for some word w1 over A( κ (i, j)−). 134 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Consider each of these cases.

−→ −1 −→ −→ −1 Case 1. W contains κ (i, j,ω) w1 κ (i, j,ω)w κ (i, j,ω) for some w. −→ ←− Case 1.1. Let j>1.Then κ (i, j)− = κ (i, j − 1).Sinceforeveryt ≥ 0 if c is −→ ←− ←− t active for κ (i, j)-sectors then it locks κ (i, j − 1)-sectors, ct locks κ (i, j − 1)-sectors. This contradicts Lemma 6.9. −→ Case 1.2. Let j = 1.Then κ (i, j)− ≡ ρ(i). Then by Lemma 6.9, t, t + 1 = 0. Recall also that t, t+1 = 2n+1 and that all other (working) rules except for rules from −1 M0∪{c0} are right active for ρ(i)-sectors. By Lemma 6.10 (ii) h and ghg are a powers of rootω(w1) modulo Burnside relations, so by Lemma 4.19, h and g commute modulo Burnside relations, a contradiction.

←− −→ −→ − Case 2. W contains κ (i, j − 1,ω)w κ (i, j,ω)w κ (i, j,ω) 1, j>1.Thenc ←− 1 t locks κ (i, j − 1)-sectors (indeed, every connecting rule which does not lock some −→ ←− κ (s, l)-sectors locks all κ (s, l)-sectors). Lemma 6.28 now implies that rules of one ←− of the sets Mt or Mt+1 lock κ (i, j − 1)-sectors. Let Mt , t ∈{t, t + 1} consist of rules ←− which lock κ (i, j −1)-sectors. By deﬁnition of M, t ∈{0, 2j −1, 2j, 2j +1, ..., 2n+1}. −→ Since for t ∈{2j,...,2n + 1} the rules of Mt lock also κ (i, j)-sectors, we conclude that t ∈{0, 2j − 1}.

Case 2.1. Let t = 2j − 1, j>1.Thent = t + 1 because rules from M lock −→ −→ ←−2j κ (i, j)-sectors. So t = 2j − 2.SinceallrulesfromM − lock all κ (i, 1)-, κ (i, 1)- ←− −→ 2j←−1 ←− ,..., κ (i, j − 1)-sectors, none of the letters κ (i, 1,ω), κ (i, 1,ω), ..., κ (i, j − 1,ω) belong to the sector W (by Lemma 6.9). The deﬁnition of admissible words now im- ←− pliesthatinW, K-letters to the left of κ (i, j − 1,ω) are placed in the natural (de- ←− −→ creasing) order from κ (i, j − 1,ω) to κ (i, 1,ω).TheK-letter in W next to the left −→ −→ − of κ (i, 1,ω) may be either ρ(i) or κ (i, 1,ω) 1. Consider both cases.

−→ Case 2.1.1. Suppose that W contains a sector of the form ρ(i)w κ (i, 1,ω). Since neither t nor t + 1 are equal to 0 or 2n + 1 (recall that t ∈{t, t + 1}, t = 2j − 1, j>1), all rules from M ∪ M + ∪{c } are right active for the ρ(i)-sectors and −→ t t 1 t (by Lemma 6.4) La(h, κ (i, 1)) is a copy of h.SinceW · h = W, we conclude that h = 1 modulo Burnside relations, a contradiction.

Case 2.1.2. Suppose that W contains a sector of the form −→ − −→ κ (i, 1,ω) 1w κ (i, 1,ω). Then by Lemma 6.10 h and ghg−1 belong to the same cyclic subgroup generated by rootω(w ) modulo Burnside relations. As before, by Lemma 4.19, this implies that h and g commute modulo Burnside relations, a contradiction. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 135

= −1 ∪ ∪{ , −1} Case 2.2. Let t 0.Thenghg contain rules from M0 M1 c0 c0 only. −1 Notice that g cannot contain a subword of the form c0g1c0 where g1 is a word over M1. Indeed otherwise by Lemma 6.11 applied to the sector ←− −→ κ (i, j − 1,ω)w1 κ (i, j,ω) of W,wegetg1 = 1 modulo Burnside relations (since c0 locks this sector, and all rules from M1 are left active for it). This would contradict Lemma 6.26. Similarly h cannot −1 contain a subword of the form c0h1c0 where h1 consists of rules from M1.Hence −1 , either h contains rules from only one Mω or h has the form h1c0 h2c0h3 where h1 h3 are possibly empty words over M1, h2 is a word over M0. Since W and W · g are in the domain of h, W and W · g have the same Ω- coordinate. Hence the total degree of c0 in g is 0. Therefore either g does not contain −1 −1 , c0 and c0 or g has the form g1c0 g2c0g3 where g1 g3 are words over M1 (these words can be empty) and g2 is a non-empty word over M0. Consider these two cases separately.

−1 −1 Case 2.2.1. Suppose that g does not contain c0 and c0 .Sinceghg must con- −1 −1 , tain c0 or c0 , h has the form h1c0 h2c0h3 where h1 h3 are words over M1, h2 is a word over M .SinceW · h = W,theΩ-coordinate of W is 1,andg is a word over M . 0 ←− −→ 1 Let us apply Lemma 6.4 (i) to the sector κ (i, j −1,ω)w1 κ (i, j,ω) of W.Since a ←− a ←− c0 locksthissector,R (gh1, κ (i, j − 1)) = R (h1, κ (i, j − 1)) = w1 modulo Burnside relations. By Lemma 6.4 (iii), we obtain that gh1 = h1 modulo Burnside relations, so g = 1 modulo Burnside relations, a contradiction.

≡ −1 , Case 2.2.2. Suppose that g g1c0 g2c0g3 where g1 g3 are words over M1, g2 is awordoverM0. If h consists of rules from M1, then we can apply Lemma 6.11 to the subword −1 −1 [←κ−( , − )−→κ ( , )] · −1 g3hg3 of ghg and a sector of the type i j 1 i j of W g1c0 g2c0 to obtain −1 = that g3hg3 is equal to 1 modulo Burnside relations. Hence h 1 modulo Burnside relations, a contradiction. ¯ −1 −1 Since h does not have subwords of the form c0hc0 , h has the form h1c0 h2c0h3 where h1, h3 are words over M1, h2 is a word over M0. Then again by Lemma 6.11, = −1 = = −1 g3h1 1, h3g3 1 modulo Burnside relations. Hence h1 h3 modulo Burnside relations, which contradicts Lemma 6.27 if h1 or h3 are not empty or Lemma 6.24 if h1 and h3 are empty.

Remark 6.30. — Notice that the only properties of the sector ←− −→ κ (i, j − 1,ω)w1 κ (i, j,ω) used in Case 2.2 are the properties that all rules from M0 ∪{c0} lock this sector and all rules from M1 are active for this sector. 136 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

−→ −→ −→ −1 Case 3. Let W contain sectors ρ(i)w1 κ (i, 1,ω) and κ (i, 1,ω)w κ (i, 1,ω) . Recall that t, t + 1 = 2n + 1.

Case 3.1. Suppose that t = 0. Then all rules from Mt ∪ Mt+1 ∪{ct} are right active for ρ(i)-sectors. Therefore by Lemma 6.12 h = 1 modulo Burnside relations, a contradiction. Case 3.2. Let t = 0. By Remark 6.30, this case is similar to Case 2.2 because all rules from M0 ∪{c0} lock ρ(i)-sectors and all rules from M1 are (right) active for these sectors. ←− ←− − Lemma 6.31. — W cannot contain a sector of the form [ κ (i, j) κ (i, j) 1], i = 1,...,N, j = 1,...,n. Proof. — Suppose that W contains a sector W of the form ←− ←− − κ (i, j,ω)w κ (i, j,ω) 1, i = 1, ..., N, j = 1, ..., n,ω∈ Ω. As in the proof of Lemma 6.29, we can assume that W contains a sector W of the form [z−1z] to the left of W , z ∈ K. We also assume that W is the closest to W such ←− sector (in principle, W can have a common letter κ (i, j,ω) with W ). ←− Since all rules from M0 ∪ M2n+1 ∪{c2i, i = 0, ..., n} lock κ (i, j)-sectors, none of these rules can occur in ghg−1 by Lemma 6.9. Therefore ghg−1 involves rules from at most two sets Mt, Mt+1. −1 If ghg contains rules of only one set Mω then by Lemma 6.13 (h, g) is not a counterexample to part (2) of Proposition 6.17, a contradiction. Thus ghg−1 contains c for some t.Moreoverthist must be odd because con- t ←− necting rules with even indices lock κ (i, j)-sectors. −→ Then c locks κ (k,)-sectors, 1 ≤ k ≤ N, 1 ≤ ≤ n. So by Lemma 6.9, ←− t z ≡ κ (k,) for any k,. ←− Hence the ﬁrst letter κ (i, j,ω) of W does not belong to W . Therefore W con- −→ ←− tains a sector of the form [ κ (i, j) κ (i, j)].IfrulesfromM and M + are active for −→ t t 1 κ (i, j)-sectors, and c locks these sectors, we get a contradiction with Lemma 6.28. t −→ Hence rules from Mt+1 must lock κ (i, j)-sectors. Therefore these rules lock all −→ ←− −→ κ (i, j )-and κ (i, j )-sectors with j

Proof. — Suppose that each sector of W has one of these forms. But for each

z ∈{λ(i,ω),λ(i,ω)−1, ρ(i), ρ(i)−1},

La(τ, z) = Ra(τ, z) = 1 for every τ ∈ M.NoticethatW is in the domain of g because ghg−1 is reduced and W · ghg−1 = W. Now by Lemma 6.4 we have W · g = W, a contradiction.

Lemma 6.33. — W does not contain sectors of the form [κ(i)κ(i)−1].

Proof. — If i>1then the statement immediately follows from Lemma 6.9 because all rules in M lock κ(i)-sectors. So let i = 1 and suppose that W contains a sector of the form [κ(1)κ(1)−1]. Then by Lemma 6.9 ghg−1 does not contain rules from ±1, ±1 M0 and M2n+1 and connecting rules c0 c2n . As before we may assume that there is −1 −1 asectorW of the form z w1z to the left of W in W (if not we replace W by W ). In particular, W is not a preﬁx of W.TheK-letter preceding κ(1) of W in W is either κ(1)−1 or λ(1,ω). In the ﬁrst case we get a contradiction using Lemma 6.10 (ii) −1 −1 and Lemma 6.19 (iii) because by Lemma 6.4 (iii) ϑ1(ghg ) is a copy of ghg in this case. In the second case we get a contradiction with Lemma 6.7.

− −→ Lemma 6.34. — W does not contain sectors of the form z 1wz,wherez ≡ κ (i, j,ω) ←− for j = 1 or z ≡ κ (i, j,ω) or z ≡ λ(i,ω).

Proof. — Indeed, suppose that W contains such a sector W .ThenbyLem- −1 −1 −→ ma 6.9, rules from M + ∪{c , c } do not occur in ghg if z ≡ κ (i, j,ω) for ←− 2n 1 2n 2n j = 1 or z ≡ κ (i, j,ω) or z ≡ λ(i,ω). Since W starts and ends with the same K-letter, we can assume that W contains −1 asectorW of the form z1w1z1 to the right of W . By Lemmas 6.29 and 6.31 the K- letter z1 in W can be of one of the forms λ(s,ω),κ(s), ρ(s) for some s. By Lemma 6.9 −1 this implies that rules from M0 and c0 cannot occur in ghg .AsintheproofsofLem- −1 ∪ ∪{ , −1} = , mas 6.29, 6.31, all rules in ghg are contained in Mt Mt+1 ct ct , t 0 2n.

−→ Case 1. Suppose that the K-letter in W is κ (i, j,ω). Then by Lemma 6.29 the −→ − ←− K-letter next to the right of W cannot be κ (i, j,ω) 1,soitmustbe κ (i, j,ω).But −→ rules from Mt ∪ Mt+1 are active for κ (i, j)-sectors and the rule ct locks this sector. Thus we get a contradiction with Lemma 6.28. ←− Case 2. Suppose now that z ≡ κ (i, j,ω). Then by Lemma 6.31, the K-letter ←− − −→ next to the right of W cannot be κ (i, j,ω) 1,soitmustbe κ (i, j + 1,ω) if j

Case 3. Suppose that the K-letter in W is λ(i,ω). Since by Lemma 6.32 not all sectors of W have the form zwz−1 where z ∈{λ(i,ω),λ(i,ω)−1}, W either has a sector − ←− of the form [λ(i)κ(i)] or it has sectors of the forms [λ(i)λ(i) 1] and [ κ (i −1, n)λ(i)]. Case 3.1. Suppose that W has a sector of the form [λ(i)κ(i)]. Recall that rules −1 from M0, c0, c2n do not occur in ghg . All other rules are right active for λ(i)-sectors. This by Lemma 6.7 implies that ghg−1 = 1 modulo Burnside relations, a contradiction. Case 3.2. Suppose that W has a sector of the form [λ(i)λ(i)−1] and a sector of ←− the form [ κ (i−1, n)λ(i)]. Since we have already proved (Cases 1, 2) that W does not ←− − ←− −→ − −→ contain sectors of the form [ κ (p, q) 1 κ (p, q)], [ κ (p, q) 1 κ (p, q)], q = 1, W must −→ ←− ←− −→ contain sectors of the forms, [ κ (i − 1, n) κ (i − 1, n)], [ κ (i − 1, n − 1) κ (i − 1, n)], −→ ←− −→ − ...,[ κ (i − 1, 1) κ (i − 1, 1)].ThusW contains κ (i − 1, 1,ω) 1.TheK-letter next to −→ − −→ − the right of κ (i − 1, 1,ω) 1 in W is either κ (i − 1, 1,ω) or ρ(i − 1) 1.Sinceall −→ −1 rules from Mt ∪ Mt+1 ∪{ct} are left active for κ (i − 1, 1) -sectors, in the ﬁrst case we get a contradiction by using Lemma 6.10(ii) and Lemma 4.19. In the second case we get a contradiction with Lemma 6.7.

−→ ←− Lemma 6.35. —IfW contains a sector W of the form [ κ (i, n) κ (i, n)] andasector [←κ−( , )λ( + )] ±1 ∓1 W of the form i n i 1 ,andh contains a subword h of the form cs h1cs ,whereh1 does not contain either connecting rules or rules from M2n+1,thenh1 contains rules from M0.

Proof. — Indeed suppose that h1 does not contain rules from M0.Noticethatcs locks either W or W , but all rules from M\(M ∪M + ∪{c ,...,c }) are left active −→ ←− 0 2n 1 0 2n for κ (i, n)-sectors and for κ (i, n)-sectors. Hence we can apply Lemma 6.11 to either

W or W , h1 is equal to 1 modulo Burnside relations which contradicts Lemma 6.26. −→ ←− Lemma 6.36. —IfW contains a sector of the form [ κ (i, n) κ (i, n)] andasectorof ←− the form [ κ (i, n)λ(i + 1)] then h contains a rule from M0 ∪{c0} or h is a word over M2n+1. Proof. — Let ω be the Ω-coordinate of W. Suppose h does not contain rules from M0 and does not consist of rules from M2n+1. Then by Lemma 6.35, h has the form ... −1 ... −1 hωcωhω+1 h2nc2nh2n+1c2n h2n cω hω where h and h are words over M , j = ω, ..., 2n + 1. j j j −→ ←− The connecting rule cω locks either κ (i, n)-sectors or κ (i, n)-sectors. The rule −→ ←− c locks both κ (i, n)-sectors and κ (i, n)-sectors. All rules from Mω, ω <2n,areleft 2n ←− −→ active for κ (i, n)-sectors, and for κ (i, n)-sectors, rules from M2n are left active for ←− −1 κ (i, n)-sectors. Hence by Lemma 6.4(iii) hω = (hω) modulo Burnside relations. By

Lemma 6.27, hω and hω are empty. Then h is not cyclically reduced (it starts with cω −1 and ends with cω , a contradiction). NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 139 −→ ←− Lemma 6.37. — Suppose W contains a sector of the form [ κ (i, n) κ (i, n)] and a sector ←− −1 of the form [ κ (i, n)λ(i + 1)].Thenghg contains rules from M0.

−1 Proof. — Suppose by contradiction that ghg doesnotcontainrulesfromM0. By Lemma 6.36 h is a word over M2n+1. Since W · h = W,theΩ-coordinate of W is 2n + 1. Suppose that g consists of rules from M2n+1. Suppose that rules from M2n+1 are active for a sector of the form [zz ] of W.Thenz ≡ κ(1), ρ(1)−1.HencebyLem- ma 6.13, either g and h commute modulo Burnside relations or W · g = W,acon- tradiction. If rules from M2n+1 are not active for sectors of W then W · g = W by Lemma 6.4(i), a contradiction. Hence g contains some rules not from M2n+1. Since W is in the domain of g (indeed W is in the domain of ghg−1 and ghg−1 is ∪{ −1} reduced), g must start with a rule from M2n+1 c2n .Sinceg contains some rules not ± ∓ 1 1 from M2n+1, g has a subword of the form ct f ct where f is a word over some Mt , t = 0. By Lemma 6.11, g is not Burnside-reduced, a contradiction with Lemma 6.26. −→ ←− Lemma 6.38. —IfW contains an occurrence of κ (i, j,ω) or κ (i, j,ω) or λ(i,ω) −1 then ghg contains rules from M0.

Proof. — Consider the cyclic graph labeled by Λ(ω) where ω is the Ω-coordinate of W. We can consider this graph as an “inverse automaton”, so that we can read clockwise and counterclockwise. Of course when we pass an edge labeled by z in the counterclockwise direction, we read z−1,whenwepassitintheclockwisedirection, we read z. Then the word z1z2...z1 of K-letters from W can be read on this automaton, and the start edge coincides with the end edge. This interpretation makes −→ ←− it clear that if W contains κ (i, j,ω), κ (i, j,ω) or λ(i,ω) and does not contain sec- − −→ ←− ←− − − tors of the form [zz 1] with z ∈{κ (p, q,ω), κ (p, q,ω), κ (p, q,ω) 1, λ(p,ω) 1} for any p, q (see Lemmas 6.29, 6.31, 6.34) then W must contain a sector of the form −→ ←− ←− [ κ (p, n) κ (p, n)] and a sector of the form [ κ (p, n)λ(p + 1, n)]. Now the statement of the lemma follows from Lemma 6.37.

Lemma 6.39. — W does not contain sectors of the form zwz−1 where z ∈ K.

Proof. — Suppose that W contains such a sector. By Lemmas 6.29, 6.31, 6.33, z ∈{λ(i,ω),ρ(i)}.

Case 1. Let z ≡ λ(i,ω). Since by Lemma 6.34 W does not contain sectors of − ←− the form [λ(i) 1λ(i)], W must contain a sector of the form [ κ (i − 1, n)λ(i)].Since ←− − ←− by Lemma 6.34 W does not contain sectors of the form [ κ (i − 1, n) 1 κ (i − 1, n)], −→ ←− W must contain a sector of the form [ κ (i − 1, n) κ (i − 1, n)]. Now by Lemma 6.37, 140 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

−1 ghg contains a rule from M0. But this possibility cannot occur because rules from M0 lock λ(i)-sectors.

−1 Case 2. Let z ≡ ρ(i).Thenghg does not contain rules from M0 by Lem- ma 6.9. −→ ←− Suppose that W contains a K-letter of the form κ (i, j,ω), κ (i, j,ω)or λ(i,ω). Then as in the proof of Lemma 6.38, by Lemmas 6.29, 6.31, 6.34 W must contain −→ ←− ←− sectors of the form [ κ (i, n) κ (i, n)] and [ κ (i, n)λ(i)]. But this contradicts Lem- ma 6.37. Therefore all K-letters in W belong to the set {κ(i), κ(i)−1, ρ(i), ρ(i)−1}.ByLem- ma 6.32 not all K-letters belong to the set {ρ(i), ρ(i)−1}. By Lemma 6.33, not all of them belong to {κ(i), κ(i)−1}. Therefore for some i from 1 to N each sector of W has one of the four forms [κ(i)ρ(i)], [κ(i)−1κ(i)], [ρ(i)ρ(i)−1], [ρ(i)−1ρ(i)] (recall that sectors of the form [κ(i)κ(i)−1] cannot appear by Lemma 6.33). Moreover a sector of the form [κ(i)−1κ(i)] must appear among sectors of W (the K-letter preceding κ(i) in W cannot be λ(i,ω) since W does not contain letters of the form λ( j,ω)). If rules from M2n+1 do not appear in g, h then we can consider any sector of W of the form [κ(i)−1κ(i)], and use Lemma 6.10 to get a contradiction. Therefore rules −1 −1 from M2n+1 do appear in ghg . By the same reason not all rules in ghg are from −1 M2n+1. Therefore ghg contains c2n.Sincec2n locks κ(i)-sectors, sectors of the form [ρ(i)−1ρ(i)] do not appear in W. Therefore for some i all sectors of W have the form [κ(i)−1κ(i)], [κ(i)ρ(i)] or [ρ(i)ρ(i)−1]. But this contradicts the assumption of part (ii) of Proposition 6.17.

Recall that W is an admissible word which starts and ends with the same letter and does not contain accepted subwords. By replacing W with W−1 if necessary we can assume that the ﬁrst letter in W belongs to K(ω) for some ω ∈ Ω.ThenbyLem- ma 6.39 all K-letters in W belong to K(ω).SinceW is an admissible word, it must contain an admissible subword W whose projections onto K-letters is equal to a cyclic shift of Λ(ω).

Lemma 6.40. — W is an accepted word.

Proof. — By definition, W contains sectors of the form [zz+] for every z ∈ K(ω). −→ ←− ←− In particular, W contains sectors of the form [ κ (1, n) κ (1, n)] and [ κ (1, n)λ(1, n)]. −1 −1 By Lemma 6.37, ghg contains a rule from M0. By Lemma 6.13, ghg contains rules −1 ±1 from at least two different Mi.Henceghg also contains either rules from S1 or c0 . −1 −1 −1 Letusprovethatghg or gh g contains a subword c0h1c1 where h1 is a word −1 ∪{ , −1}∪ over S1. Suppose that it is not true. Then ghg contains rules from M0 c0 c0 M1 −1 −1 Ω · · only. Hence ghg contains c0 and c0 .Sincethe -coordinates of W, W g, W gh are the same, the total exponent of c0 in g is 0 and the total exponent of c0 in h is 0. −1 If g or h contains a subword c0 f c0 where f is a word over M1, then by Lemma 6.7, NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 141 ←− applied to any κ (i, j)-sector of W,wegetthatf = 1 modulo Burnside relations, a contradiction with Lemma 6.26. Hence neither g nor h contain subwords of the −1 form c0 f c0 . Now consider two cases. ω = · −1 = ±1 Case 1. Suppose that 0.SinceW ghg W, the first occurrence of c0 −1 −1 in ghg is c0.Sinceh does not contain subwords of the form c0 f c0 ,thisc0 occurs · Ω −1 in g.ButsinceW g has the same -coordinate as W, g must contain c0 ,henceg −1 contains a subword of the form c0 f c0 , a contradiction. ω = ±1 −1 −1 Case 2. Suppose that 1. Then the first occurrence of c0 in ghg is c0 . Since W · g = W, we can represent g as g1g2 where g2 is the maximal suffix of g consisting of rules from M1 and g1 is either empty or ends with c0.SinceW · h = W, we can represent h as h1h2h3 where h1 and h3 consist of rules from M1 and either − ←− h = h is empty or h starts with c 1 and ends with c .Considerany κ (1, 1)-sector 2 ←−3 −→ 2 0 0 U ≡ κ (1, 1, 1)u κ (1, 2, 1) of W.

−1 Case 2.1. Suppose that g1 is empty, g ≡ g2 consists of rules from M1.Sinceghg contains c , h must contain c ,soh is not empty. 0 0 2 ←− Since c0 locks U and rules from g are left active for κ (1, 1)-sectors, by Lem- a a −1 −1 ma 6.4 (i), we have R (gh1) = u = (R (h3g )) modulo Burnside relations. By Lem- = −1 = −1 ma 6.4 (iii), we have that gh1 gh3 modulo Burnside relations. Hence h1 h3 modulo Burnside relations. This contradicts Lemma 6.27 if h1 or h3 is not empty or Lemma 6.24 if h1, h3 are empty. Case 2.2. Suppose that g is not empty. Then g ends with c−1.Ifh does not 1 1 0 −→ contain c0,thatish2, h3 are empty, then by Lemma 6.7 applied to a κ (1, 1)-sector −1 = = of W, g2hg2 1 modulo Burnside relations, so h 1 modulo Burnside relations, a contradiction. If h contains c0,thatish2 is not empty, then again by Lemma 6.7 g2h1 = 1, −1 = = −1 h3g2 1 modulo Burnside relations. Thus h1 h3 modulo Burnside relations which as before contradicts Lemma 6.27 or Lemma 6.24.

Thus, since one can always consider h−1 instead of h, we can assume that h has the form ec0h1c1 f for some words e, f .Sincec0 locks z-sectors, z ∈{λ(i), κ(i), ρ(i), ←− κ (i, j)}, a cyclic shift of W · e has the form −→ ←− −→ ←− λ(1, 0)κ(1)ρ(1) κ (1, 1, 0)u1,1 κ (1, 1, 0) κ (1, 2, 0)u1,2 κ (1, 2, 0)... −→ ←− κ (N, n, 0)uN,n κ (N, n, 0).

The word W · ec0 has the form −→ ←− −→ ←− λ(1, 1)κ(1)ρ(1) κ (1, 1, 1)u1,1 κ (1, 1, 1) κ (1, 2, 1)u1,2 κ (1, 2, 1)... −→ ←− κ (N, n, 1)uN,n κ (N, n, 1). 142 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR −→ Since c1 locks κ (i, j)-sectors, for every i ∈{1, ..., N} and every j ∈{1, ..., n},wehave

a −→ R (h1, κ (i, j)) = ui, j modulo Burnside relations. Hence all ui, j are equal modulo Burnside relations, so acyclicshiftofW · e is equal to the word −→ ←− −→ ←− λ(1, 0)κ(1, 0)ρ(1, 0) κ (1, 1, 0)u κ (1, 1, 0) κ (1, 2, 0)u κ (1, 2, 0)... (22) −→ ←− κ (N, n, 0)u κ (N, n, 0) for some word u which is equal to ui, j modulo Burnside relations. Denote by u¯ the copy of u written in M0. Then by Lemma 6.4 W · eu¯ = Λ(0),soW is accepted.

Lemma 6.40 gives us the ﬁnal contradiction because by our assumption W cannot contain an accepted subword. Proposition 6.17 is proved.

7. Bands, annuli, and trapezia

7.1. Deﬁnitions and basic facts

∗ In Section 6, we introduced an auxiliary group Hka whichwasthefreeproduct of the subgroup K freely generated by K and the subgroup Ha generated by the set A subject to the R(κ(1))-relations and the Burnside relations. Recall that the subgroup Ha is the free product in the variety of Burnside groups of exponent n of its subgroups A(z), z ∈ K. From now on let us call all relations of Ha as a-relations.All relations from Z(M) involving letters from K will be called κ-relations and all relations involving letters from R will be called r-relations. We also will call the commutativity relations from Z(M) involving R-andA-letters ra-relations. Accordingly we shall consider a-cells, κ-cells, r-cells and ra-cells in van Kampen diagrams. Notice that κ-cells are also r-cells. Recall that the set of relations Z(M, Λ) consists of Z(M) and the hub Λ(0). The cells corresponding the hub relation, will be called hubs as well. In this section, we shall consider diagrams over the set Z(M, a) of all a-relations and all r-relations (including κ-relations). With every word over K ∪ R ∪ A we associate its K-length, its R-length and its A-lengthinthenaturalway.InanydiagramoverZ(M, a) we consider R-edges and A-edges as 0-edges and K-edges as non-zero edges. In a diagram is over the presentation consisting of a-relations and ra-relations, then we consider R-edges as non-zero edges and A-edges as zero edges. Thus the metric on a diagram depends on over what presentation this diagram is. Let S be one of the sets K, R, A. Then we can consider S-bands in any diagram over Z(M, a) as in Section 2.2. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 143

• K-bands consist of κ-cells. A maximal K-band which is not an annulus can start and end on the boundary of a diagram. All K-letters of a K-band belong to the −1 same set K(z) for some z ∈ K ∪K . Thus we can consider K(z)-bands as well. • R-bands consist of r-cells. A maximal R-band which is not an annulus must start and end on the boundary of the diagram. For every R-band T there exists a rule τ ∈ M such that R-edges in an R-band have labels of the form r(τ, z), z ∈ K. Thus we will sometimes call T a τ-band. The next lemma immediately follows from the definition of words L(. , z) and R(. , z) and the definition of K-relations. Lemma 7.1. — Consider a disc or an annular diagram consisting of one K(z)-band T . Then for some word h over M (which is called the history of the band), the word written on the top of T is R(h, z), the word written on the bottom of T is L(h, z).ThewordR(h, z) (resp. L(h, z))isR-reduced if and only if T does not contain two consecutive cells that cancel. Moreover L(h, z)z = zR(h, z) modulo κ-relations, where z, z are the labels of the start and end edges of the band. The next lemma contains easy properties of R-bands which follow immediately from the definition of Z(M) and Lemma 5.1. Lemma 7.2. —Let

w ≡ u0z1u1z2...zsus be the bottom side of a τ-band T , ≡ ... w u0z1u1z2 zs us T , ∈ ∪ ( )−1 , ≡ ... be the top side of ,wherezi zi K K ui ui are words over A.ThenW z1u1z2 zs ≡ ... · τ = and W z1u1z2 zs are admissible words and W W .

7.2. Forbidden annuli Let ∆ be any (disc or annular) diagram over Z(M, a).Wecall∆ Z(M, a)-reduced if every K-band is reduced (that is they do not contain consecutive K-cells that cancel) and every disc subdiagram whose boundary label is a word over A and is equal to 1 modulo a-relations does not contain R-edges. Lemma 7.3. —Let∆ be a Z(M, a)-reduced disc diagram. Then ∆ does not contain 1. K-annuli, 2. (K, R)-annuli, 3. R-annuli, 4. disc subdiagrams with some r-cells whose boundary path contains no R-edges, 144 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Proof. — We are proving these statements by a joint induction on the number of cells of the inside diagrams of the annulus or the “bad” subdiagrams. By contradiction take the smallest (with respect to the number of cells) disc subdiagram ∆ of ∆ which either contains one of the annuli forbidden by the lemma or has boundary path forbidden by the lemma.

Case 1. Assume there is a K-annulus Q in ∆. Then there exists an R-edge on asideofQ.HenceQ crosses an R-band T .SinceQ and T form a (K, R)-annulus with the inside diagram smaller than that for Q, we obtain a contradiction.

Case 2. Suppose that ∆ is the inside diagram of a (K, R)-annulus W1 ∪ W2 where W1 is a K-band and W2 is an R-band. Then the contour of ∆ does not contain K-edges (otherwise there would be a (K, R)-annulus with a smaller inside diagram). Therefore W1 consists of two cells. These two cells have a common K-edge and they belong to the same r-band W2. Therefore these two κ-cells cancel, a contradiction.

Case 3. Suppose that there is a κ-cell in an R-annulus Q.Thenwegetacon- tradiction as in Case 1. Therefore the boundary label W of ∆ is a word in A.Since the counterexample is minimal, ∆ has no edges except for A-edges. Hence W = 1 modulo a-relations. Diagram ∆ has the same boundary label W,sinceall(r, a)-cells in Q, are commutativity cells. But this contradicts the property that ∆ is Z(M, a)- reduced, because there are R-edges in Q.

Case 4. Suppose that ∂(∆ ) contains no R-edges. Then by part 3 of the lemma, ∆ contains no R-edges, so it cannot contain r-cells, a contradiction. ∗ ( , ) Corollary 7.4. —ThegroupHka is (naturally) embedded into the group deﬁned by Z M a .

Proof. — Indeed, let W be a word in K∪A which is equal to 1 modulo Z(M, a). Then there exists a Z(M, a)-reduced van Kampen disc diagram ∆ with boundary label W. By Lemma 7.3, part 4, ∆ contains no r-cells, so all cells in ∆ are a-cells. = = ∗ ∗ Hence W 1 modulo a-relations, so W 1 in Hka. Thus the natural map of Hka into the group given by the relations Z(M, a) is an embedding.

7.3. Trapezia Here we shall present a tool to translate results from Section 6 into the language of van Kampen diagrams and vice versa. Let ∆ be a Z(M, a)-reduced disc diagram which has the contour of the form −1 −1 p1q1p2 q2 where: NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 145

(TR1) φ(q1), φ(q2) are K-reduced words without R-letters, starting and ending with K-letters, and (TR2) p1 and p2 are sides of reduced K-bands each containing at least one κ-cell.

Then ∆ is called a trapezium.Thepathq1 is called the bottom base,thepathq2 is called the top base of the trapezium, the paths p1 and p2 are called the left and right sides of the trapezium. The history of the K-band whose side is p1 is called the history of the trapezium.

By Lemma 5.1, if a word h over M contains just one rule τ, W ≡ z1u1...zt is a reduced admissible word, and W · h = W1, then there exists a word W which is τ · = equal to W modulo a-relations and such that is applicable to W , W h W1 where Θ W1 is equal to W1 modulo a-relations. By Lemma 5.1 there exists a trapezium con- τ = sisting of one -band with bottom base label W , top base label W1.SinceW W , = ∆, ∆ W1 W1 modulo Burnside relations, there exist diagrams 1 consisting of a-cells ( )−1 ( )−1 ∆ with boundary labels W W and W1 W1 respectively. Gluing three diagrams , Θ, ∆1 in a natural way, we get a trapezium T(W,τ) with bottom base label W,top base label W1, left side label L(τ, z1), right side label R(τ, zt ). If a reduced word h consists of several rules, the corresponding trapezium can be obtained by induction. Let h = h τ where h is a word of smaller length than h. Suppose that we have constructed a trapezia T(W, h ) with the bottom base label W and the top base label W·h , and suppose that W·h exists. Then W·h is in the domain of τ,soτ is applicable to a reduced admissible word W1 which is equal to W·h mod- ∆ −1 ulo a-relations. Then there exists a diagram over the a-relations with contour s1s2 where φ(s1) = W · h ,φ(s2) = W1. By Lemma 5.1 there exists a trapezium T(W1,τ) with the bottom base label W1.GlueT(W, h ) with the diagram ∆ identifying the top base of T(W, h ) with s1. Then glue the resulting diagram with T(W1,τ) identifying s2 with the bottom base of T(W1,τ).TheK-bands in the resulting diagram T(W, h) are reduced because h is a reduced word. Every closed path in T(W, h) consisting of A-edges is contained in one of the subdiagrams ∆ consisting of a-cells. Hence T(W, h) is a Z(M, a)-reduced diagram. It is clear that conditions (TR1) and (TR2) are satis- ﬁed. Hence T(W, h) is a trapezium with bottom base label W, top base label W · h, left side label L(h, z1) and right side label R(h, zt).Ithasaformofasandwichwhere diagrams consisting of a-cells are sandwiched between τ-bands (see Figure 14). Clearly T(W, h) is not determined uniquely by W and h because we can choose the a-parts of the sandwich in many different ways. Any trapezium of the form T(W, h) is called a normal trapezium with history h and base label W.

Lemma 7.5. — Every trapezium is a normal trapezium T(W, h) for some W and some h.

Proof. — Consider a trapezium ∆ with bases q1, q2 and sides p1, p2. By condition (TR2) p1 and p2 contain R-edges. Since by (TR2), p1 and p2 are sides of reduced K- 146 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR · z1 W h zt " " " $ $ $ $ $ %% %% $ $ $ $ $ " " "" " % % $ $ % % $ $ " " $ &"%$" % % $ $ % % $ $ % % # R(h, zt) L(h, z1) $ $ % ! !" $ " % % $ $ & % % $ $ & % % $ $ ' % % $$ $$ ' % % z1 W zt

FIG. 14.

bands, φ(p1), φ(p2) are R-reduced. Since ∆ is Z(M, a)-reduced, we can apply Lem- ma 7.3(2) and conclude that every maximal R-band starting on p1 ends on p2 and every maximal R-band starting on p2 ends on p1. By Lemma 7.3(3), ∆ does not contain any other maximal R-bands.

Since by (TR1) φ(q1) and φ(q2) are K-reduced, by Lemma 3.2 every maximal K-band starting on q1 (resp. q2)endsonq2 (resp. q1) and by Lemma 7.3(1) ∆ does not have any other maximal K-bands.

Let T , T be the ﬁrst and the last maximal K-bands in ∆ starting on q1 counting from p1 to p2.LetB and B be the ﬁrst and the last R-bands of ∆,countingfrom q1 to q2.

Consider the subdiagram ∆ of ∆ bounded by q1 and the connecting line of B. −1 By Lemma 7.2 the boundary label of ∆ is equal to φ(q1)W where W is an admissible word. Since ∆ does not contain r-cells (by Lemma 7.3, part 3), W = φ(q1) ∗ φ( ) φ( ) in Hka.Since q1 is reduced by (TR1), W is a reduced admissible word and q1 is a reduced admissible word which is equal to W.Let∆ be the diagram obtained ∆ ∆ B ∂(∆ ) = ( )−1( )−1 from by removing and .Then p1q1 p2 q2 where p1 is a side T T B of without the ﬁrst cell, p2 is a side of without the ﬁrst cell, q1 is a side of , = φ( ) = φ( ) · τ ∗ B τ q2 q2. Since by Lemma 7.2 q1 q1 in Hka if is a -band, we have that φ( ) q1 is K-reduced.

Suppose that p1 does not have R-edges. Then p2 does not have R-edges either (every maximal R-band of ∆ is a maximal R-band of ∆), so the boundary of ∆ contains no R-edges. By Lemma 7.3 (4) ∆ does not contain r-cells, hence φ(q2) = φ( ) = φ( ) · τ ∗ ∆ = (φ( ), τ) q1 q1 in Hka,hence T q1 as required. ∆ Now if p1 contains R-edges then p2 also contains R-edges, so satisﬁes conditions (TR1) and (TR2), so ∆ is a trapezium. Since ∆ contains fewer maximal R- ∆ ∆ = (φ( ), ) bands than , by induction, we can assume that T q1 h for some h.Then ∆ = T(φ(q1), τh) as required NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 147

8. The group Hra

Consider the group Hra which is the group generated by A ∪ R subject to the set of relations consisted of all a-relations and all (commutativity) ra-relations from Z(M, a).WeconsiderA-letters as zero-letters and R-letters as non-zero letters. So in terminology of Section 3, in this section Y = R. We are going to prove that the presentation of Hra satisfies conditions (Z1), (Z2), (Z3) from Section 3 Locally, in this section, we are going to keep the same notation as in Section 3. We let R0 = S0 be the set of a-relations and ra-relations. The set S1/2 is thus empty. Clearly conditions (Z1.1) and (Z1.2) hold. Since S1/2 is empty, conditions (Z2.1), (Z2.2) and (Z2.3) also hold. We will use the terminology and results from Section 3.1 and Section 3.2. In particular, we are going to talk about essential, reduced and cyclically reduced words in Hra. As in Section 3, for every cyclically reduced essential element g ∈ Hra we define 0(g) = 0(g) as the maximal subgroup of Ha which contains g in its normalizer. For an arbitrary essential element g of the form vuv−1 where u is cyclically reduced, we define 0(g) = v0(u)v−1. By Lemma 3.7, 0(g) is well defined (does not depend on the presentation g = vuv−1). The following lemma will imply both (Z3.1) and (Z3.2).

Lemma 8.1. —Letg be an essential element represented by a cyclically reduced word A,let x = 1 be a 0-element such that g−4xg4 is a 0-element. Then the following two statements hold. (i) There exists a 0-element u such that gu−1 commutes with every element of 0(g). (ii) x ∈ 0(g).

Proof. — By Lemma 3.6, g−2xg2 is 0-element. Let B be a word in A representing a non-trivial element x from 0(g), C be a word in A representing g−2xg2. Let A ≡ u1r1u2...rsus+1 where ri ∈ R, i = 1, ..., s, ui are words in A, i = 1, ..., s+1. By Lemma 3.7 (all of the conditions of that lemma hold), we can replace A by any cyclic shift of A. So we can assume that u1 is empty. Then there exists a diagram ∆ of minimal type over the presentation S0 of Hra −1 −1 φ( ) ≡ φ( ) ≡ 2 φ( ) ≡ φ( ) ≡ with boundary p1q1p2 q2 where q1 q2 A , p1 B, p2 C. Since A is cyclically reduced and p1, p2 do not contain R-edges, every maximal R-band in ∆ starting on q1 ends on q2 and every maximal R-band starting on q2 ends on q1. By Lemma 7.3 (3), ∆ contains no other maximal R-bands. Let T1, ..., T2s be all maximal R-bands starting on q1, counted from p1 to p2. Notice that every ra-relation is a commutativity relation of the form ra = ar where r ∈ R(z), a ∈ A(z) for some z ∈ K.Foreveryi = 1,...,2s, ri belongs to R(zi) for some zi ∈ K. Then the labels of the top and the bottom paths of Ti are equal to 148 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

−1 awordvi in the alphabet A(zi) ∪ A(zi) .NoticethatB isconjugateofeachofthe words vi, i = 1,...,2s in Hra. Indeed, the subdiagram of ∆ bounded by p1, q1, q2 and T −1 −1 = the connecting line of i has boundary label t Btvi for some word t.SinceB 1 modulo in Hra,noneofthewordsvi is equal to 1 modulo in Hra. The connecting lines of the bands T1, ..., T2s divide ∆ into diagrams ∆0,...,∆2s+1 such that the boundary of each ∆i contains no R-edges. By Lemma 7.3 (4), each ∆i has no r-cells, so the boundary label of each ∆i is equal to 1 in the group Ha. Since A starts with r1, the subdiagram ∆0 is bounded by a side of T1 and p1. Hence B = φ(p1) = v1 modulo a-relations. ∆ −1 −1 −1 = The boundary label of 1 is u2 v1u2v2 .Henceu2 v1u2 v2 in Ha.Recallthat Ha is the free product of subgroups A(z) in the variety of Burnside groups of exponent n. By Lemma 36.2, 34.11 from [30], each of these subgroups is malnormal in Ha. Therefore v2, u2 ∈A(z1) modulo a-relations. Since u2, v2 = 1, z1 ≡ z2.Now considering subdiagrams ∆2, ..., ∆s+1 one by one, we conclude that ui, vi are words in A(z1), i = 1,...,s + 1,andz1 = z2 = ... = zs. Therefore all ri belong to R(z1).Since vi are not empty, ri commutes with some letters from A(zi). By deﬁnition of Z(M) then ri commutes with every letter from A(z1) in Hra.Hencer1, ..., rs commute with all letters from u1,...,us+1.Inparticular,A = r1r2...rsu1...us+1 in Hra. Let u = u1...us. This word does not contain letters from R.NoticethatHa is aretractofHra, u is the Ha-projection of g,sou does not depend in Hra on the word A representing g.Letalsog1 = r1...rs.SinceR is a retract of Hra, g1 does not depend −1 on the word A representing g, gu = g1. We have proved that B is equal modulo a-relations to a word in A(z1).Sog1 commutes with B = x.Since0(g) is a 0-subgroup normalized by g, it consists of 0- −2 2 elements y such that g yg is a 0-element. We have proved that g1 commutes with every such y.Hencegu−1 commutes with all elements in 0(g) which proves (i). Notice that the subgroup A(z1) is normalized by g = g1u because g1 commutes with this subgroup and u belongs to it. Hence A(z1)⊆0(g).Sincex ∈A(z1), x ∈ 0(g), which proves (ii).

We have checked all properties (Z1), (Z2), (Z3), so by Proposition 3.19, we have the following statement.

Proposition 8.2. — There exists a graded presentation Rra of the factor group Hra(∞) of Hra over the subgroup generated by all n-th powers of elements of Hra, such that every g-reduced diagram over Rra satisﬁes property A from Section 2.4, and all the lemmas from Section 3 hold.

As a corollary of Proposition 8.2, we have the following statements which we are going to use later.

∆ R ∂(∆) = −1 −1 Lemma 8.3. —Let be a g-reduced diagram over ra, p1q1s q2 , obtained by gluing two diagrams Γ and Π where NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 149 ∂(Γ) = −1 −1 φ( ) ( , ) (i) p1q1p2 q2 , p1 is a subword of R h z for some reduced word h over M −1 and some z ∈ K ∪ K , p1 starts and ends with R-edges; q1 and q2 do not contain R-edges; Γ is a diagram of rank 0 (that is over Z(M, a)); n −1 (ii) Π is a cell of corresponding to a relation v = 1 in Rra of rank ≥ 1, ∂(Π) = p2s ; l (iii) v starts with an R-letter r, φ(p2) ≡ v r where l>1/2εn.

Then there exists a diagram ∆ over Rra which has the same boundary label as ∆, ∂(∆ ) = ( )−1( )−1 φ( ) ≡ φ( ), φ( ) ≡ φ( ) φ( ) ≡ φ( ), φ( ) ≡ φ( ) p1q1 s q2 where p1 p1 q1 q1 , s s q2 q2 ,and (1) ∆ contains a subdiagram Π where φ(∂(Π )) ≡ un where u is a cyclically R-reduced word, |∂(Π ) ∩ | | | (2) p1 >6u , (3) ∆ \Π is of rank 0.

Proof. — Consider the R-bands in Γ starting on p2.Sincev is a period, it is cyclically R-reduced (by Proposition 8.2 and the deﬁnition of periods from Section 3. Since q1 and q2 are of R-length 0, every R-band starting on p2 ends on p1.Sinceφ(p1) is a subword of an R-reduced word R(h, z),everyR-bands starting on p1 ends on p2. Hence the R-projections of words φ(p1) and φ(p2) are identical (the labels of R-edges in all cells of a R-band are the same). l 2 Therefore φ(p1) ≡ u r where |u|R =|v|R, u starts with r.Sinceu and u are subwords of an R-reduced word R(h, z) (since l ≥ εn>2), u is cyclically R-reduced. In particular, u is an essential word.

Let T and T be the ﬁrst and the last maximal R-bands in Γ starting on p1 (we count the R-bands from (p1)− to (p1)+). Consider the subdiagram Γ1 of Γ bounded by the top side of T , p1, p2 and the connecting line of T (so this subdiagram is equal to Γ minus T ). Since all cells in T are commutativity cells, x1 = φ(top(T )) =

φ(bot(T )). The subdiagram of Γ bounded by a side of T and q1 does not contain R-bands, so φ(q1) = x1 modulo a-relations. Similarly φ(q2) = x2 modulo a-relations where x2 is the word written in a side of T . Clearly |x1|R =|x2|R = 0. ¯ ¯ Let p1 be the path p1 without the last edge, p2 be p2 without the last edge. Then ¯ l ¯ l φ(p1) ≡ u , φ(p2) ≡ v ,

l l (23) u x1 = x2v modulo a-relations and ra-relations. Similarly every subword of the cyclic word u is equal to a subword of the same R-length of the cyclic word v up to multiples of zero length. Hence v is cyclically reduced by Lemma 3.3. Therefore v is essential. Suppose that u is not simple in rank 0. By Lemma 3.25 there exists a word w which is simple in rank 0 (and so w is simple in rank 1/2 because Rra does not b −1 contain hubs), |w|R < |u|R,andawordt ∈ 0(w) such that u = xw tx in rank 0, 150 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR for some word x and some integer b = 0.Sinceu is cyclically reduced, |x|R = 0 and b |w |R =|u|R. l bl −1 Therefore by u = xw t1x for some t1 ∈ 0(w) (since 0(w) is normalized by w). Using (23), we get

l bl −1 l u x1 = xw t1x x1 = x2v in rank 0, where both w and v are simple in rank 0, the R-lengths of x, x1, t1, x2 are equal to 0 (for x, x1, x2 we have established it above, for t1, this follows from the deﬁn- ition of 0(w) since w is cyclically R-reduced). Now we can apply Lemma 3.32 and ±1 −1 conclude that v = yw t2 y in rank 0 for some word y and some t2 ∈ 0(w).This contradicts the fact that v is simple in rank 0 (since |w|R < |u|R =|v|R). Thus u is simple in rank 0 (and so it is simple in rank 1/2). Then again by Lemma 3.32, we can conclude that ±1 −1 = (24) x2v t2x2 u in rank 0 for some t2 ∈ 0(v). Suppose that the exponent of v in (24) −1,thatis −1 −1 = x2v t2x2 u. Therefore l = ( −1 )l −1 = −l −1 u x2 v t2 x2 x2v t x2 for some t ∈ 0(v) (since 0(v) is normalized by v). By (23), −1 l = l −1 = −l −1 x2 u v x1 v t x2 whence 2l = −1 . v t x2 x1 Since t is of R-length 0 (this follows from the deﬁnition of 0(v) since v is cyclically 2l R-reduced), x1 and x2 are also of R-length 0,wehavethatv is of R-length 0 which contradicts the fact that v is cyclically R-reduced. This contradiction shows that in fact the exponent of v in (24) is 1. ( )n = n n = n −1 By Lemma 3.12, vt2 v in rank 0. Hence u x2v x2 in rank 0. φ( ¯ ) ≡ l ∆¯ ¯ ( )−1 Recall that p1 u .Let be a diagram with contour p1 p1 where φ( ) ≡ l Π Π¯ p1 u , which is a result of gluing two mirror copies and with boundary la- ±n R Π ¯ bels u over ra where p1 is a part of the boundary of , p1 is a part of the boundary Π ∆ ∆¯ | |≥ | |≥ 1 ε | | | | of . Consider the union of and .Noticethat p1 l u 2 n u >6u .Itre- mains to observe that in the union of ∆ and ∆¯ ,theΠ¯ is connected with Π by a path with label x2, hence this pair of subdiagrams and the connecting path can be replaced n = n −1 by a diagram of rank 0 corresponding to the equality u x2v x2 . The lemma is proved. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 151

9. The group Hkra

Recall that in Section 7 we considered the set of deﬁning relations Z(M, a) which is the union of the set of all a-relations and the set Z(M) of relations corresponding to the S-machine M.

Now consider the union Z(M, Λ, ra) = Z(M, a) ∪{Λ(0)}∪Rra and the group Hkra generated by K ∪ R ∪ A subject to the relations from Z(M, Λ, ra). Each relation of Z(M, Λ, ra) is either the hub, an a-relation, a (commutativity) ra-relation, or a κ-relation or belongs to Rra. Burnside relations from Rra will be called Burnside ra- relations. The corresponding cells in diagrams will be called Burnside ra-cells.Thesetof all relations from Z(M, Λ, ra) except the hub will be denoted by Z(M, ra).

As in Section 3, we divide the set of generators of Hkra into the set of 0-letters and the set of non-0-letters. Our main goal is to prove that the presentation Z(M, Λ, ra) satisfies properties (Z1), (Z2), (Z3) if we consider K as the set of non-zero letters and the R ∪ A as the set of 0-letters. ButatfirstweshowhowtogetridofBurnsidera-cells in diagrams over Z(M, ra). As in Section 7, we can consider K-andR-bands in diagrams over Z(M, ra). The definition of these bands is the same as in Section 7 (Burnside cells do not belong to R-bands). The only difference is that now R-bands can start and end on Burnside ra-cells.

9.1. Getting rid of Burnside ra-cells

Consider (temporarily) R as the set of non-0-letters and all other letters as 0- letters. We can consider R-bands connecting two Burnside ra-cells (a Burnside ra-cell and a part of the boundary of a diagram) as 0-bonds, and deﬁne bonds and contiguity subdiagrams of all ranks. (We preserve the grading of the set of Burnside ra-relations Rra.) Suppose that a disc van Kampen diagram ∆ is obtained by gluing together two disc diagrams Θ and Π over Z(M, ra) along pieces of their boundaries, where Θ ( , ) −1 −1 , (1) is a trapezium T W h with contour p1q1p2 q2 (q1 q2 are the bottom and the top bases); (2) φ(Π) ≡ un for some cyclically R-reduced word u over R ∪ A,thediagramΠ does not contain K-edges;

(3) Π and Θ are glued along a subpath p of p2; |p|≥3|u|. Then we shall say that ∆ has the form Θ ∗ Π. Suppose that a disc van Kampen diagram ∆ is obtained by gluing together two diagrams Π and Θ where 152 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR Θ ( , ) −1 −1 , (1) is a trapezium T W h with contour p1q1p2 q2 (q1 q2 are the bases); (2) φ(Π) ≡ un for some cyclically R-reduced word u over R ∪ A containing R- letters; Π does not contain K-edges; (3) Π and Θ are glued along a subpath p of p1. Then we shall say that ∆ has the form Π ◦ Θ.Itiseasytoseethat T(W, h) ∗ Π = Π ◦ T(W · h, h−1) (this means that if the right hand side of the equality exists then both sides exist and are equal).

Lemma 9.1. — For every van Kampen disc diagram ∆ of the form Θ ∗ Π there exists a van Kampen disc diagram ∆ of the form Π ◦ Θ with the same boundary label and such that: (i) the label of the bottom (top) base of Θ is equal to the label of the bottom (top) base of Θ ; (ii) The histories of the trapezia Θ and Θ are equal modulo Burnside relations (see Fig- ure 15).

W · h

· W h ' q2

−→ Π Θ p1 p Π Θ ( p2

q1 W W

FIG. 15.

−1 −1 Proof. — Let Θ = T(W, h).Thenφ(p2) ≡ R(h, z) where z ∈ K ∪ K is the last letter of W.Letp be the common subpath of the contour of Π and the right side of the trapezium Θ, φ(∂(Π)) ≡ un for a cyclically R-reduced word u.Thenφ(p) ≡ R(h, z)−1 contains a cube of a cyclic shift of u as a subword. Without loss of generality we can assume that this cyclic shift is u itself. The word R(h, z)−1 is a product of blocks of the form R(τ, z), τ ∈ M, each of which is either a one-letter word r ∈ R ∪ R−1 or a two-letter word (ra)±1 where r ∈ R, a ∈ A ∪ A−1. Every block is completely determined by its R-letter. Consider the occurrence of u3 in R(h, z)−1. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 153

The word u3 contains subword v2 where v is a cyclic shift of u and the ﬁrst occurrence of v (in v2) starts with the beginning of a block. Suppose that v does not end with the end of a block. Then the second occurrence of v in v2 starts in the middle of a block, so the ﬁrst letter of v is both a beginning of a block and an end of a block. ±1 −1 −1 Therefore this letter belongs to A .Hencev ≡ ar1v r2 where a ∈ A ∪ A , r1 ∈ R , r2 ∈ R, ar1 and r2a are blocks. But then the second letter of the second occurrence of v must be r1, and it must be the beginning of a block. Thus the same R-letter r1 is a beginning and the end of a block. This contradicts the fact that every block is completely determined by its R-letter. Hence v starts with a beginning of a block and ends with the end of a block. ≡ ( , ) 2 Therefore v R h1 z for some cyclically reduced word h1 over M.Thush1 is a sub- ≡ 2 , · 2 word of h.Leth h h1h for some words h h .ThenW h is in the domain of h1. · s By Lemma 6.5(iv), then W h is in the domain of h1 for every integer s.Inparticular, · 2−n W h is in the domain of h1 . By Lemma 6.5(ii), · = · 2 = · 2−n . W h W h h1h W h h1 h

n Since v ≡ R(h1, z) is a cyclic shift of u,onecanreadv on the boundary of Π, and the boundary label of Θ ∗ Π is freely equal as a cyclic word to WR(h , z)−1vn−2R(h , z)−1(W · h)−1L(h, z ) = ( , )−1 ( 2−n, )−1 ( , )−1( · )−1 ( , ) = (25) WR h z R h1 z R h z W h L h z ( 2−n , )−1( · )−1 ( 2 , ) WR h h1 h z W h L h h1h z where z is the ﬁrst letter of W. Θ = ( , 2−n ) Consider now the trapezium T W h h1 h which exists by Lemma 7.5(i) 2−n · 2−n = · since W is in the domain of h h1 h .SinceW h h1 h W h, we can assume that the top base label of Θ is the same as the top base label of Θ,thatisW · h. The boundary label of Θ is thus freely equal (as a cyclic word) to ( 2−n , )−1( · )−1 ( 2−n , ). WR h h1 h z W h L h h1 h z Notice that by Proposition 8.2 there exists a van Kampen disc diagram Π con- n taining only a-cells, ra-cells and Burnside ra-cells, that has boundary label L(h1, z ) . 2−n We can glue Π and Θ along the path labeled by L(h1, z ) ,wegetadiagram∆ of the form Π ◦ Θ . The boundary label of this diagram will be freely equal to ( 2−n , )−1( · )−1 ( 2 , ). WR h h1 h z W h L h h1h z Comparing this with (25) we conclude that the boundaries of ∆ and ∆ are freely equal. The labels of the bottom (top) bases of Θ and Θ are equal by construc- ≡ 2 Θ 2−n Θ tion, the history h h h1h of is equal to the history h h h of modulo Burnside relations, as required. 154 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Lemma 9.1 allows us to move subdiagrams with boundary labels of the form un from one side of a trapezium to another side, preserving the history of the trapezium modulo Burnside relations. The next lemma is the ﬁrst application of this trick.

Lemma 9.2. — For every disc diagram ∆ over Z(M, ra) there exists another disc diagram ∆ with the same boundary label and without K-annuli.

Proof. — We can assume that ∆ contains exactly one K-annulus B and ∂(∆) is equal to t = top(B). Then the subdiagram ∆ bounded by bot(B) is a diagram over Rra. We can assume that this subdiagram is g-reduced, and that the K-annulus B is reduced. By Proposition 8.2 ∆ is an A-map. Suppose that ∆ contains a Burnside ra-cell. By Lemma 2.6 there exists a Burnside ra-cell Π in ∆ and its contiguity subdiagram Γ to bot(B) with contiguity degree at least ε. By Lemma 8.3, we can replace Π ∪ Γ by a new subdiagram Ψ with the same boundary label which contains a subdiagram Π such that φ(∂(Π )) ≡ un for some word u and |∂(Π ) ∩ bot(B)| >6|u|, Ψ\Π is a diagram of rank 0. Now using Lemma 9.1, we can move Π through B without producing new K-annuli. After a number of such transformations we replace ∆ by a new disc diagram ∆¯ with the same boundary label containing exactly one K-annulus which bounds a subdiagram without Burnside ra-cells. By Lemma 7.3, part 1, applied to the subdiagram bounded by that K-band, we can replace that subdiagram by a subdiagram without K-edges. Since this operation reduces the number of K-annuli, the statement of the lemma follows.

Let ∆ be an annular diagram over Z(M, ra) with inner contour p1 and outer contour p2,andletq be a path in ∆. Suppose that ∆, q satisfy the following conditions:

(T1) p1 and p2 are sides of K-annuli T1 and T2 respectively; φ(p1) = 1 modulo Z(M, ra); (T2) q connects (p1)− = (p1)+ with (p2)− = (p2)+; the ﬁrst and the last edges of q, e1, e2,areK-edges; (T3) Every maximal K-band in ∆ is an annulus surrounding the hole of ∆. Maximal K-bands in ∆ divide ∆ into annular subdiagrams (without κ-cells) which will be called chambers. We count the chambers of ∆ from p1 to p2.

Lemma 9.3. — Suppose that all chambers of ∆ except the last one, ∆¯ ,donotcontain Burnside ra-cells and φ(q) is K-reduced. Then there exists another annular diagram ∆ over Z(M, a) (that is without Burnside ra- ∆ cells) with contours p1 and p2, and a path q in satisfying the conditions (T1), (T2), (T3) φ( ) = φ( ) φ( ) = φ( ) φ( ) = φ( ) such that p1 p1 , p2 p2 modulo Burnside relations, q q modulo Z(M, ra), φ(q ) is K-reduced. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 155

Proof. — Among all pairs (∆, q) satisfying the conditions of the lemma and having the same boundary labels of contours p1, p2 modulo Burnside relations and the same labels of q modulo Z(M, ra) letuschooseapairwhere∆ has the smallest possible number of chambers and the last chamber ∆¯ has the smallest type in the sense of Section 3. First of all let us reduce all chambers of ∆. We can get rid of j-pairs of Burnside ra-cells by ﬁrst replacing the path q by a path avoiding the j-pair and its connecting path and then reducing the j-pair. Similarly, if ∆ contains a disc subdiagram Ψ over Z(M, a) containing r-cells, whose boundary label is a word over A and is equal to 1 modulo a-relations, then we can replace this subdiagram by a subdiagram with the same boundary label containing no r-cells. The resulting diagram will satisfy conditions (T1), (T2), (T3). These operations will not increase the number of chambers in ∆ or the type of ∆¯ or the K-length of φ(q). Thus we can assume that ∆ does not contain disc subdiagrams whose boundary labels are words over A equal 1 modulo a-relations. In particular, all chambers except for the last one are Z(M, a)-reduced in the sense of Section 7.2.

We can also reduce all K-annuli in ∆ and also the bands T1 and T2 without changing the K-length of q. Notice that we do not assume that T1 and T2 are cyclically reduced because it may be impossible to reduce the annuli T1 and T2 without changing q. Hence we can assume that ∆¯ is g-reduced in the sense of Section 3 as a diagram ¯ over Z(M, ra) and ∆\(∆ ∪ T2 ∪ T1) which is a diagram over Z(M, a) is Z(M, a)- reduced.

Since φ(q) is K-reduced in Hkra, q intersects every K-annulus in ∆ only once. Therefore the K-length of q is equal to the number of K-annuli in ∆. ¯ Let B be a maximal R-band connecting the contours of ∆ = ∆\(∆ ∪ T2). Such a band exists because if every R-band starting on the boundary of T1 ends on the boundary of T1 then the label of a side of T1 is equal to 1 modulo Z(M, ra), a contradiction with (T1). Since the label of a side of B is equal to the label of the portion of q in that subdiagram multiplied by words of K-length 0 (read along the boundaries of the subdiagram), the label of any side of B is K-reduced. ¯ By Proposition 8.2, chamber ∆ is an A-map. Let o = (e2)−.Lets be the inner contour of ∆¯ ,andlett be the outer contour of ∆¯ .Weconsidero as the beginning andtheendoft. Suppose that ∆¯ contains Burnside ra-cells. Then by Lemma 2.5, part c), there exists a Burnside ra-cell Π in ∆¯ and its contiguity subdiagram Γ ofrank0tooneof the contours s or t of contiguity degree ≥ ε. Replacing Γ (if necessary) by a smaller contiguity diagram, we can assume that conditions (i)–(iv) of Lemma 8.3 hold. By this lemma, we can replace Γ∪Π by a subdi- 156 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR agram Ψ with the same boundary label, which has a subdiagram Π such that ∂(Π ) ≡ un where u is a cyclically R-reduced word and |∂(Π ) ∩ s| >6|u| or |∂(Π ) ∩ t| >6|u| and Ψ\Π has rank 0.

Suppose first that |∂(Π )∩t| >6|u|. By Lemma 9.1 we can move Π through T2 replacing T2 by a K-band with a history which is equal to the history of T2 modulo Burnside relations. The path q can be replaced in the new annular diagram by a path with the same label modulo Z(M, ra). This operation decreases the type of the last chamber of ∆. Suppose now that |∂(Π ) ∩ s| >6|u|.Letx = ∂(Π ) ∩ s.Considertheannular subdiagram ∆ bounded by s and p1.Thediagram∆ is a Z(M, a)-reduced diagram. By Lemma 7.3 every maximal R-band starting on x must end on p1. Indeed otherwise the R-band would have two intersections with a reduced K-band in a disc subdiagram of ∆ which contradicts Lemma 7.3, part 2. Since such an R-band cannot intersect twice a K-annulus, and φ(q) is a K-reduced word, the label of a side of any of these R-bands is a K-reduced word. There exists a disc subdiagram Θ of ∆ containing at least half of these R-bands, bounded by a subpath y of x, a subpath of p1 and by two r-bands. Now Θ is a trapezium because it satisfies conditions (TR1) and (TR2). By Lemma 7.5, Θ is a normal trapezium. The union of Θ and Π is, by definition, equal to Θ∗Π .AsbeforewecanmoveΠ through Θ replacing Θ by a trapezium with the same history modulo Burnside relations. Again we decrease the type of ∆. The new diagram satisfies conditions (T1), (T2), (T3) and has a smaller type than ∆. Thus we can complete the proof by induction on the type of ∆.

Lemma 9.3 implies the following stronger statement.

Lemma 9.4. — The conclusion of Lemma 9.3 holds if one removes the restriction that only the last chamber contains Burnside ra-cells.

Proof. — Since φ(q) is reduced, we can assume that it crosses the connecting line of every K-annulus only once. Let us use induction on the number of chambers in ∆.Ifthenumberofcham- ¯ bers is 1, we can apply Lemma 9.3. Consider the diagram ∆1 = ∆\(∆ ∪ T2) where ¯ ∆ is the last chamber. Let q1 be the part of q in ∆1.Thenumberofchambersin∆1 is smaller than the number of chambers in ∆. Therefore we can apply the induction hypothesis and conclude that there exists a diagram ∆2 and a path q2 which satisfy the conclusion of Lemma 9.3.

Since the label of the outer contour of ∆2 is equal to the label of the outer contour of ∆1 modulo Burnside relations, we can glue to the outer contour of ∆2 adiagramoverRra such that the resulting diagram ∆3 has the same outer boundary label as ∆1. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 157

Now replace the subdiagram ∆1 in ∆ by ∆3 and the subpath q1 of q by q2.Let (∆4, q4) be the resulting pair of an annular diagram and a cutting path. Reducing if necessary, the last K-annulus of subdiagram ∆3 of ∆4, we obtain a pair (∆5, q5) satisfying conditions of Lemma 9.3. Applying that lemma, we complete the proof. The following lemma is an algebraic consequence of Lemma 9.4.

Lemma 9.5. —LetP1, P2 be words over A∪R which are not equal to 1 modulo Z(M, ra). −1 Let A P1A = P2 in Hkra for some word A which is K-reduced in Hkra and starts and ends with −1 a letter z ∈ K ∪ K .ThenA = R1BR2 in Hkra where R1 = L(h1, z), R2 = R(h2, z), , −1 z is in the domain of h1 h2 , B is a reduced admissible word starting and ending with z1 such that R1z1 = zR(h1, z), z1R2 = L(h2, z1)z modulo Z(M, a).

−1 Proof. — Equality P2 = A P1A gives us an annular diagram ∆ with inner contour p1 labeled by P1,outercontourp2 labeled by P2 and a cutting path q labeled by A. The connecting line of any K-annulus in ∆ not surrounding the hole of ∆ cannot cross q because otherwise by Lemma 9.2 we would be able to decrease the K- length of q by going along the boundary of such a K-annulus instead of crossing it. Hence by Lemma 9.2 we can assume that ∆ does not contain K-annuli not surrounding the hole of ∆. Let e and f be the first and the last edges of q.Sincee and f are K-edges, they belong to the first and the last K-annulus of ∆ counting from p1 to p2.Thelabelof ( , ) ∆ p1 is equal modulo Z M ra to the label of a side p1 of the first K-annulus in and the label of p2 is equal in Z(M, ra) to the label of a side of the last K-annulus in ∆. Hence we can assume that p1 and p2 are the sides of these annuli. Thus ∆ satisfies conditions (T1), (T2), (T3). Since φ(q) is K-reduced, we can apply Lemma 9.4 and conclude that there exists another annular diagram ∆ over ( , ) ∆ Z M a with contours p1 and p2,andapathq in satisfying the conditions (T1), φ( ) = φ( ) φ( ) = φ( ) φ( ) = (T2), (T3) such that p1 p1 , p2 p2 modulo Burnside relations, q φ(q) = A modulo Z(M, ra), φ(q ) is K-reduced. Assume that every maximal R-band starting on the first K-annulus of ∆ does ( , ), not end on p2. Then the label of p1 is equal to 1 modulo Z M ra a contradiction. ∆ B Hence contains an R-band , starting on p1 and ending on p2. B ( )s( ) Let t be a side of .Thenq is homotopic to p1 p1 tp2 for some integer s, φ( ) ≡ where p1 is a subpath of p1 and p2 is a subpath of p2. By Lemma 7.2 t B is an admissible word starting and ending with a K-letter z .TheK-length of B is the same as the K-length of A (since B intersects each of the K-annuli exactly once), whence B is reduced in Hkra. φ(( )s( ) The word p1 p1 is written on the top side of z -band, so it is equal to ≡ ( , ) φ( ) ≡ ( , ) R1 L h1 z for some h1. Similarly p2 R h2 z for some h2. The last equalities from the lemma follow from Lemma 7.1. 158 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Lemma 9.6. —LetA be a K-reduced word without R-letters, starting and ending with −1 letters z1, z2 from K ∪ K , and suppose that for some words P1 and P2 of K-length 0, P1 = 1 −1 modulo Z(M, ra), we have A P1A = P2 modulo Z(M, ra).Then:

(1) P1 = L(h, z1), P2 = R(h, z2) modulo Z(M, ra) for some word h over M; (2) A is a reduced admissible word; (3) A · h = A.

Proof. — Consider a van Kampen disc diagram ∆ over Z(M, ra) for the equality −1 = ∂(∆) = −1 −1 φ( ) ≡ −1,φ( ) ≡ = , A P1A P2, p1q1p2 q2 where pi Pi qi A, i 1 2.ByLem- ma 9.2 we can assume that ∆ does not contain K-annuli. We can also assume that all K-bands in ∆ are reduced. Hence every maximal K-band in ∆ connects q1 and q2. These maximal K-bands divide ∆ into diagrams over Z(M, ra) without K-edges, which we shall call chambers as in the annular case before. As in the proof of Lem- ma 9.4, one can use induction on the number of chambers, Proposition 8.2, Lem- ma 8.3 and Lemma 9.1, to move all Burnside ra-cells out of ∆. Since only two sides of ∆,namelyp1 and p2 may contain R-edges, only these two sides will change during this process. Their labels will stay the same modulo Z(M, ra). Hence there exists a van Kampen diagram ∆1 over Z(M, a) (i.e. over Z(M, ra) but without Burnside ra-cells) ( )−1 −1 φ( ) = φ( ) = , with contour p1q1 p2 q2 where pi pi , modulo Burnside relations i 1 2. Let T1 and T2 be the ﬁrst and the last K-bands in ∆1 starting on q1 and end- ∆ ∆ ing on q2 (we count the K-bands starting at p1). Consider the subdiagram 2 of 1 T T T , T ∂(∆ ) = ( )−1 −1 bounded by 1, 2, q1, q2 (and containing 1 2). Then 2 p1q1 p2 q2 T T where p1 is a side of 1, p2 is a side of 2. ∆ ( )−1 ( )−1 Considering the subdiagrams of 1 bounded by p1 p1 and p2 p2 ,wecon- φ( ) = φ( ) φ( ) = φ( ) ( , ) clude that p1 p1 , p2 p2 modulo Z M ra . This gives part (1) of the lemma.

The subdiagram ∆2 is a trapezium (by deﬁnition). Hence it is a normal trapezium T(A, h) for some word h (by Lemma 7.5). In particular, A is an admissible word φ( ) = φ( ) = φ( ) = ( , )−1 which proves part (2) of the lemma. Moreover, p1 p1 p1 L h z1 , φ( ) = φ( ) = φ( ) = ( , )−1 ( , ) p2 p2 p2 R h z2 modulo Z M ra for some word h over M.Since the top base of ∆2 has label A,wehaveA · h = A. This gives part (3) of the lemma.

Lemma 9.7. —LetB be a K-reduced word, starting with a letter z ∈ K∪K−1 and ending with a letter z ∈ K ∪ K−1 which does not contain R-letters. Let U be a word over A,letP, R be words of K-length 0 such that BURBUz is K-reduced, P = 1 in Z(M, ra). Suppose that (BURBUz)−1P(BURBUz) is a 0-word Q modulo Z(M, ra).ThenBUz is an admissible word, and there exist 0-words P , R and Q which are equal modulo Z(M, ra) respectively to P, R, Q , such that (BUR BUz)−1P (BUR BUz) = Q modulo Z(M, a). NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 159

Proof. — Consider a disc diagram ∆ over Z(M, ra) corresponding to the equality

(BURBUz)−1P(BURBUz) = Q ,

∂(∆) = −1 −1 φ( ) ≡ φ( ) ≡ φ( ) ≡ φ( ) ≡ p1q1p2 q2 where p1 P, p2 Q , q1 q2 BURBUz.Let = = φ( ) ≡ φ( ) ≡ φ( ) ≡ φ( ) ≡ q1 s1x1t1s1, q2 s2x2t2s2 where si B, xi U, ti R, si BUz, i = 1, 2.LetT1, T2 be the first and the last K-bands starting on s1, T3, T4 be the T , T first and the last K-bands starting on s1.SinceBUR is cyclically reduced, 1 2 end T , T on the first and the last edges of s2, 3 4 end on the first and the last edges of s2. ¯ Then we can assume that p1 is a side of T1, p2 is a side of T4.Letp1 be the side of T ( ) ( ) ¯ T ( ) 2 which connects s2 + with s1 +,letp2 be a side of 3 which connects s2 − with ( ) s1 −. By Lemma 9.6 φ(p1) = L(h1, z), φ(p2) = R(h2, z) modulo Z(M, ra), B and BUz areadmissiblewords,andB · h1 = B, BUz · h2 = BUz. Thus as in Lemma 9.4 we can move all Burnside cells out of the subdiagrams ∆1 and ∆2 of ∆ bounded by ( ¯ )−1 −1 ¯ −1( )−1 ( , ) ( , ) p1s1 p1 s2 and by p2s1p2 s2 , and replace them by trapezia T B h1 and T B h2 in the expense of possible increasing the number of cells in the subdiagram Γ bounded ¯ ¯−1( )−1 ∆ = ( , ) ∆ = ( , ) by p1x1t1p2 x2t2 . Hence we can assume that 1 T B h1 , 2 T B h2 . Thus all Burnside ra-cells in ∆ are in Γ. Consider the diagram Γ.ByLem- ma 9.2 we can assume that Γ does not have K-annuli. Since the boundary of Γ does not contain K-edges, the entire diagram Γ does not have K-edges. ¯ Since φ(t1) ≡ φ(t2),wecanidentifyt1 and t2 to obtain an annular diagram Γ −1 ¯ ¯ Γ and a path t connecting the contours x2 p1x1 and p2 of . We can eliminate all j- pairs of Γ¯ ,replacingt by a path with the same label modulo Z(M, ra). Hence we can assume that Γ¯ does not have j-pairs. Hence Γ¯ is an A-map by Proposition 8.2. If Γ¯ does not have Burnside ra-cells, we are done, because we can again cut Γ¯ along the path t. Suppose that Γ¯ contains Burnside ra-cells. Consider Γ¯ as a map satis- ¯ ¯ fying condition A from Section 2.4 with boundary divided into four parts: x1, x2, p1, p2. Recall that paths x1, x2 do not contain R-edges. By Lemma 2.5, there exists a Burnside ¯ ¯ ra-cell Π and a contiguity subdiagram Γ of rank 0 of Π to p1 or p2 with contiguity degree at least ε. We can assume that t avoids Γ .HenceΠ and Γ are subdiagrams of the (disc) diagram Γ.AsintheproofsofLemmas9.3,9.4,wecanuseLemmas8.3 and 9.1 and move Π through either the trapezium ∆1 or the trapezium ∆2 leaving φ(p1) and φ(p2) the same modulo Z(M, ra),leavingpathssi, xi untouched and de- creasing the number of Burnside ra-cells in Γ. The proof now can be completed by induction on the number of Burnside ra-cells in Γ.

9.2. Conditions (Z1), (Z2), (Z3) Now we are ready to check that the presentation Z(M, Λ, ra) satisﬁes conditions (Z1), (Z2), (Z3) from Section 3. 160 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

Here we are going to use notation from Section 3 again, as in Section 8. Let R0 = S0 be the set Z(M, ra) of all relations in Z(M, Λ, ra) except for the hub, and let S1/2 ={Λ(0)},soZ(M, Λ, ra) = R1/2.WeletY = K, so all letters from R ∪ A are 0-letters. Condition (Z1.1) holds by an easy inspection. Condition (Z1.2) holds by Propo- sition 8.2. The length of Λ is (2n + 3)N>n, so (Z2.1) holds. The condition (Z2.2) is obviously true. Thenextlemmagivesus(Z2.3).

±1 Lemma 9.8. — Assume that v1w1 and v2w2 are cyclic permutations of Λ(0) and |v1|≥ ε|Λ(0)|. Then if u1v1 = v2u2 holds modulo S0 for some 0-words u1, u2, then we have u2w1 = w2u1 modulo S0.

Proof. — Suppose that u1v1 = v2u2 modulo S0. Then there exists a van Kam- Γ S −1 −1 φ( ) ≡ φ( ) ≡ pen diagram over 0 with boundary p1q1p2 q2 such that p1 u1, p2 u2, φ(q1) ≡ v1, φ(q2) ≡ v2 (see Figure 6). The K-bands of Γ starting on q1 end on q2 since −1 φ(q1) is a linear word and u1, u2 do not contain letters from K ∪ K . The labels of the start and end edges of each of these bands are equal. Hence v2 ≡ φ(q2) ≡ φ(q1) ≡ −1 v1 ≡ v.Since|v| >0,andΛ(0) and Λ(0) do not have common letters, v cannot be a subword of both Λ(0) and Λ(0)−1. Without loss of generality assume that v is a subword of Λ(0). Recall that for every j = 1, ..., N,wedenoteΛj (0) to be the subword of Λ(0) starting with λ( j) and ending with λ( j + 1) (asusual“N + 1”hereis1). By Lemma 9.6 there exists a word h over M such that v · h = v, u1 = L(h, z1), u2 = R(h, z2) modulo Z(M, ra) for some z1, z2. Hence we can assume that u1 = L(h, z1), u2 = R(h, z2) in the free group and that Γ = T(v, h).SinceT(v, h) is a diagram over Z(M, a), Γ does not contain Burnside ra-cells. Since v ≡ φ(q1) ≡ φ(q2) is a subword of Λ(0), we can attach to Γ two hubs Π1 and Π2 along the paths q1 and q2, and consider the resulting diagram ∆ with −1 −1 ∆ boundary label w2u1w1 u2 . We need to show that can be replaced by a diagram over S0 with the same boundary label. Since |v1|≥ε|Λ(0)| and N>n>3/ε, we conclude that v1 contains Λj(0) for some j>1. φ( ) ≡ Λ ( ) Λ Let q1 be the subpath of q1 such that q1 j 0 .Since is linear, every K-band starting on q1 ends on q2.LetT1 be the λ( j)-band and T2 be the λ( j + 1)- T λ( , ) T band starting on q1.Then 1 ends on the edge of q2 labeled by j 0 and 2 ends on the edge of q2 labeled by λ( j + 1, 0). T T Γ Γ The bands 1, 2,thepathq1 and the path q2 bound a subdiagram of which is a trapezium. By Lemma 7.5, Γ = T(Λj(0), h) and Λj(0) · h = Λj(0).Lett be the subpath of ∂(Γ ) which is a side of T1. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 161

Consider the subdiagram ∆ of ∆ consisting of Π1, Π2 and t.Since∆\∆ consists of 0-cells, it sufﬁces to show that the boundary label of ∆ is equal to 1 modulo S0. Thus we need to show that φ(t) = L(h, λ( j)) commutes with

−1 −1 −1 (26) Λj(0)λ( j + 1, 0) Λj+1(0)λ( j + 2, 0) ...Λj−1(0)λ( j, 0) .

By Proposition 6.16, for every i = 1,...,N we have Λi(0) · h = Λi(0).Thetrapezium T(Λi(0), h) gives us the equality

−1 Λi(0) L(h, λ(i))Λi(0) = R(h, λ(i + 1)) = λ(i + 1, 0)−1(L(h, λ(i + 1))λ(i + 1, 0), i = 1, ..., N modulo S0.Hence

−1 −1 λ(i + 1, 0)Λi(0) L(h, λ(i))Λi(0)λ(i + 1, 0) = L(h, λ(i + 1)), i = 1, ..., N. Hence if we conjugate L(h, λ(i)) by the word (26), we get L(h, λ(i)) as required.

As in Section 3, for every cyclically reduced essential element g ∈ Hkra we define 0(g) as the maximal subgroup of Hra which contains g in its normalizer. For an arbitrary essential element g of the form vuv−1 where u is cyclically reduced, we define 0(g) = v0(u)v−1. By Lemma 3.7, 0(g) is well defined (does not depend on the presentation g = vuv−1). The following lemma will imply both (Z3.1) and (Z3.2).

Lemma 9.9. —Letw be an essential element represented by a cyclically minimal in rank −4 4 1/2 word W,letx = 1 be a 0-element in Hkra such that w xw is a 0-element. Then the following two statements hold. (i) There exists a 0-element u such that wu−1 commutes with every element of 0(w). (ii) x ∈ 0(w).

Proof. — We divide the proof into several steps. 1. Let P1 be a 0-word representing x. By Lemma 3.4 every cyclic shift of W is a cyclically K-reduced word. Therefore we can assume that W starts with a letter z ∈ K ∪ K−1 (by Lemma 3.7). −1 By Lemma 3.6, (Wz) P1Wz is equal in Hkra to a 0-word P. By Lemma 9.5

Wz = CBz D in Hkra where B does not contain R-letters and starts with a K-letter z such that Cz = zC , z D = D z modulo Z(M, a) where D ≡ L( f , z), C ≡ L( f , z) for some words f , f over M. 162 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

In particular since Wz = CBz D = CBD z,wehaveW = CBD in Hkra.Since D Cz = z DC , passing if necessary to a cyclic shift BD C of W, we can assume that C is empty, so z = z,andreplaceW by BD where D z = zD modulo Z(M, a).The −1 word B is equal to zB1U where B1 is either empty or ends with a letter z ∈ K ∪ K , U is a word over A. 2. By Lemma 3.6, we have the following equalities in Hkra:

−1 (27) (zB1UD zB1Uz) P1(zB1UD zB1Uz) = P2 −1 (28) (zB1UD zB1UD zB1Uz) P1(zB1UD zB1UD zB1Uz) = P3 where P2 and P3 are 0-words.

By Lemma 9.7 (which can be applied because zB1UD zB1Uz is reduced since this word is equal to a subword of the same length of the reduced word W3)wecan assume that equality (27) holds modulo Z(M, a) (replacing P1 and P2 by equal in Hkra words if necessary), Bz is a reduced admissible word. By Lemma 3.6, since D z = zD, −1 we have that (zB1Uz) P1(zB1Uz) is a 0-word modulo Z(M, a). Hence by Lemma 9.6 P1 = L(h, z) modulo Z(M, a) and

(29) Bz · h = Bz.

3. Now consider a diagram ∆ over Z(M, a) corresponding to equality (27), ∂(∆) = −1 −1 φ( ) ≡ φ( ) ≡ φ( ) ≡ ≡ φ( ) = p1q1p2 q2 where p1 P1, p2 P2, q1 BD Bz q2 .Thenq1 φ( ) ≡ φ( ) ≡ φ( ) ≡ T s1t1s1,where s1 B, t1 D , s1 Bz. Consider the maximal z-band 1 in ∆ starting on the ﬁrst edge of s1. As usual, we can assume that p2 is a side of a z-band T2. The bands T1 and T2 bound a subdiagram ∆1 of ∆ which is a trapezium. By Lemma 7.5, ∆1 = T(Bz, h1) where h1 is the history of T1. Consider the natural homomorphism ψ from Hkra to the free group freely generated by M. This homomorphism takes K and A to 1, and takes each letter from R to the corresponding rule from M. It is clear that this homomorphism is correctly deﬁned because killing all letters from K ∪ A in every relation from Z(M, a) and replacing letters from R by the corresponding rules from M gives us a trivial relation 1 = 1.SinceD = L( f , z), ψ(D ) = f . Applying this homomorphism to the bound- −1 ary label of the subdiagram of ∆ bounded by p1 and a side of T1,wegeth1 = f hf . Hence we have

(30) Bz · f −1hf = Bz.

4. Now let ∆1 be a van Kampen diagram corresponding to equality (28). Ap- plying Lemma 9.6 to the subdiagram ∆2 of ∆1 bounded by two z-bands T1, T2 corresponding to the two last distinguished occurrences of z in zB1UD zB1UD zB1Uz, we obtain zB1Uz · h2 = zB1Uz, i.e. Bz · h2 = Bz where h2 is the history of the z- band T1. Applying the homomorphism ψ to the boundary label of the subdiagram NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 163

∆ = ∆ \∆ ψ( )−2 ψ( )2 −1 = −2 2 −1 of 2 1 2 we obtain the equality D h D h2 f hf h2 modulo Burn- −2 2 −1 = side relations. Hence f hf h2 1 modulo Burnside relations. We conclude that

(31) Bz · h2 = Bz

−2 2 where h2 = f hf modulo Burnside relations. −1 −1 Let W1 = Bz, h1 = f hf , g = f , b = h2.ThenW1·h1 = W1 by (30), W1·gh1g = −1 W1 by (29) and W1 · b = W1 by (31) where b = g h1g modulo Burnside relations. Suppose that Bz contains an accepted subword B .ThenB ·h is a cyclic shift C of Λ(0) for some word h over M. Then for some admissible subword B of B and B ,

B · h = C where C C ≡ C, |C |≤1. By Lemma 7.5 B = UC V in Hkra where U and V are 0-words. Hence B = U(C )−1V in rank 1/2.Since|U(C )−1V| < |UC V|, and B is a subword of W, we get a contradiction with the assumption that W is cyclically minimal in rank 1/2. Thus all conditions of Proposition 6.17 hold. By that proposition and by Re- mark 6.18, we can conclude that for every integer s there exists an word bs+1 over M s −s −s−1 s+1 which is equal to g h1g = f hf modulo Burnside relations and such that

Bz · bs+1 = Bz.

5. For arbitrary integer s consider the trapezia Ts = T(Bz, bs).Let∂(Ts) = −1 −1 φ( ) = ( , ) φ( ) = ( , ) φ( ) = = , p1q1p2 q2 ,where p1 L bs z , p2 R bs z , qi Bz, i 1 2. Considering the subdiagram of Ts obtained by removing the last z-band (whose side is p2), we get the equality

−1 L(bs, z)BL(bs, z) = B modulo Z(M, a).ThusB commutes with L(bs, z) modulo Z(M, a).Noticethatb0 = h. Hence B commutes with L( f , z)−sL(h, z)L( f , z)s modulo Z(M, ra).RecallthatL( f , z) −s s ≡ D , L(h, z) ≡ P1 ≡ L(b0, z).Hence(D ) x(D ) commutes with B for every integer s. Therefore for every non-negative integer s, we have the following equalities in Hkra:

−s s W P1W = −s s −1 −1 −1 −1 −1 −1 (BD ) P1(BD ) =(D ) B ...(D ) B (D ) (B P1B)D BD ...BD = −1 −1 −1 −1 −1 (32) (D ) B ...(D ) B ((D ) P1D )BD ...BD = −1 −1 −1 −2 2 (D ) B ...B ((D ) P1(D ) )B...BD = ... −s s (D ) P1(D ) (we underline the parts of the words which are changing). Similar equalities hold for −s s negative s. Therefore for every integer s, w xw is a 0-element in Hkra.Hencex ∈ 0(w) 164 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

(since the subgroup generated by w−sxws, s = 0, ±1,..., consists of 0-elements and is normalized by w). 6. Let us prove that B commutes with every element y ∈ 0(w) modulo Z(M, ra). −2 2 Let Py be a word in R ∪ A representing y.Sincey ∈ 0(w), W PyW is equal mod- −1 ulo Z(M, ra) to a 0-word. By Lemma 3.6, (BD z) Py(BD z) is a 0-word (modulo −1 Z(M, ra)). Since D z = zD modulo Z(M, ra),wehavethat(BzD) Py(BzD) is a 0- ( , ) ( )−1 ( ) ( , ) word modulo Z M ra . By Lemma 3.6, Bz Py Bz is a 0-word Py modulo Z M ra . Considering a van Kampen diagram for this equality, we deduce (by a usual argu- = ( , ) = ( , ) , ment) that Py L hy z , Py R hy z for some words hy hy over M. Applying the homomorphism ψ to the equality ( )−1 ( , )( ) = ( , ) Bz L hy z Bz R hy z = we get that hy hy modulo Burnside relations. The last equality implies −1 = −1 ( , ) = ( , ) −1 = ( , ) = (33) B PyB B L hy z B zR hy z z L hy z Py modulo Z(M, ra) (by Lemma 7.1). Hence B commutes with Py modulo Z(M, ra),as required. 7. Now let y ∈ 0(w) be represented by a word Py. Let us prove by induction on l −l l that (D ) Py(D ) belongs to 0(w) for every l ≥ 0. This is clearly true for l = 0. Suppose that it is true for some l ≥ 0.Then l+1 −l−1 l −l −1 (D ) Py(D ) = (D )(D ) Py(D ) (D ) (34) −1 l −l −1 = B (BD )(D ) Py(D ) (BD ) B. l −l l −l −1 Since (D ) Py(D ) belongs to 0(w) = 0(BD ),wehavethat(BD )(D ) Py(D ) (BD ) belongs to 0(w) (by the deﬁnition of 0(w)). Since B commutes with every element of 0(w), (34) implies that l+1 −l−1 l −l −1 (D ) Py(D ) = (BD )(D ) Py(D ) (BD ) ∈ 0(w). l −l Similarly one can prove that (D ) Py(D ) belongs to 0(w) for all l<0.

8. Let u be the element of Hkra represented by D . Then clearly u is a 0-element. Let w be the element represented by W(D )−1.Weneedtoprovethatw commutes −1 with y,thatis,W(D ) commutes with Py modulo Z(M, ra), or, equivalently, −1 −1 (BD ) Py(BD ) = (D ) Py(D ) modulo Z(M, ra). This has been proved already in (33). This completes veriﬁcation of properties (Z1), (Z2), (Z3). By Proposition 3.19, we have the following statement.

Proposition 9.10. — There exists a graded presentation Rkra containing Z(M, ra) of the factor group Hkra(∞) of Hkra over the subgroup generated by all n-th powers of elements of Hkra,such n that every relation of Rkra of rank ≥ 1 has the form u = 1 for some word u, and every minimal diagram over Rkra satisﬁes property A from Section 2.4, and all the lemmas from Section 3. NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 165

10. Proof of Theorem 1.3

Lemma 10.1. — For every word w(x1,...,xm),ifw(a1(κ(1)), ..., am(κ(1))) = 1 in Hkra(∞),thenw(a1,...,am) = 1 in A(κ(1)).

Proof. — Let ∆ be a g-reduced diagram over Rkra with boundary label w(a1(κ(1)), ..., am(κ(1))). Suppose that ∆ contains cells of rank >0.ByProp- osition 9.10 ∆ is an A-map. By Lemma 2.5 ∆ contains a cell Π of rank >0and a contiguity subdiagram of Π to ∂(∆). Since non-0 letters in presentation Rkra are letters from K, by deﬁnition of contiguity subdiagrams from Section 2, ∂(∆) contains a K-edge, a contradiction. Hence ∆ does not contain cells of rank >0. Therefore ∆ is a diagram over Hkra. So we can assume that ∆ is a g-reduced diagram over Hkra. By Lemma 9.2 we can assume that ∆ does not contain K-annuli. Hence ∆ does not contain K-edges. Therefore ∆ is a diagram over Rra where R is the set of non-0 letters. Suppose that ∆ contains a cell of rank >0. Then by Proposition 8.2 and Lem- ma 2.6 ∆ contains a cell Π of rank >0and a contiguity subdiagram of Π to ∂(∆). Hence ∂(∆) contains R-edges, a contradiction. Therefore ∆ does not contain cells of rank >0.Hence∆ is a diagram over Hra. By Lemma 7.3, part (3), ∆ does not contain R-annuli. Hence ∆ does not contain R-edges. This means that all cells of ∆ are a-cells. Therefore ∆ is a diagram over Ha.

Since A(κ(1)) is a retract of Ha,weobtainthatw(a1(κ(1)), ..., am(κ(1))) is equal to 1 modulo R(κ(1))-relations and Burnside relations, as required.

Given the Burnside group B(m, n) generated by a1, ..., am,wehave,byLem- ma 5.2, a homomorphism φ : B(m, n) → H such that φ(ai) = ai(κ(1)), i = 1, ..., m. n Assume that φ(w(a1, ..., am)) ∈ H .Thenw(a1(κ(1)), ...am(κ(1))) = 1 in n Hkra(∞) = H (n) = H /H by Proposition 9.10. Setting R(κ(1)) = ∅,wehavefrom Lemma 10.1 that w = 1 in A(κ(1))=B(a1, ..., am; n).Henceφ is an embedding, and moreover, the intersection of φ(B(m, n)) ∩ H n is trivial. This proves Statement (1) of Theorem 1.3. Let now L be an arbitrary normal subgroup of B(m, n) (the latter can now be identified with its image in H /H n). To prove Statement (2) of Theorem 1.3, we have to show that L = LH /H n ∩ B(m, n) = L. Let us define R(κ(1)) to be the set of all words in a1(κ(1)), ..., am(κ(1)) representing elements of L. The new group Hkra(∞) (whose construction depends now on the new choice of R(κ(1))) is a homomorphic image of H /H n by Proposition 9.10. Therefore it suffices to prove that every word ∼ w(a1(κ(1)), ..., am(κ(1))) that is equal to 1 in Hkra(∞),isalsotrivialinB(m, n)/L = A(κ(1)). But again, this claim is a part of Lemma 10.1. The proof of Theorem 1.3 is complete. 166 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR

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Subject index accepted admissible word 126 cyclically Y-reduced word 67 adjacent edges 56 diagram divided by a connecting line (band) 55 admissible word 108, 116, 121 domain of a word in M 117 annulus 55 0-edge 54 (S, T)-annulus 56 0-element 65 c-aperiodic word 93 essential element 65 K-band 143 graded map 54 K-band 143 graded presentation 65 R-band 143 historyof a computation 50 S-band 55 history of a K-band 143 τ-band 143 history of a trapezium 145 bases of a trapezium 145 hoop 97 bond 57 homomorphism ϑi 128 0-bond 56 homomorphism υ 127 bottom side of a band 55 inside diagram of an annulus 56 Burnside cells 98 inverse rule 109 Burnside-reduced word 120 inverse semigroup P(M) 117 Burnside ra-relations 151 Lowest Parameter Principle 46 0-cell 54 length of a path 50 chamber 154 Y-length 65 cleaning rule 111 Y-letter 65 compatibility of cells 75, 98 0-letter 54 congruence extension property (CEP) 46 map 53 connecting line of a band 55 A-map 59 connecting line of a bond 57 narrow bond 62 connecting line of a contiguity submap 59 normal trapezium 145 connecting rule 107, 111 j-pair 75 contiguity arc 59 1/2-pair 75 contiguity degree 59 period of rank i 74, 97 contiguity submap 58 A-periodic word 74 converting S-rules into sets of relations 110 phase decomposition 74 Ω-coordinate of a word 108 positive edge (cell) 54 copy of a word 118 positive and negative S-rules 110 crossing bands 55 principal cell of a bond 57 cyclically Y-reduced element 65 properties A1, A2, A3 59 168 ALEXANDER YU. OL’SHANSKII, MARK V. SAPIR properties (M1), (M2), (M3) 54 B(m, n) 46 properties (T1), (T2), (T3) 154 bot(B) 55 properties (TR1), (TR2) 144 ∂(Π1, Γ, Π2) 59 properties (Z1), (Z2), (Z3) 65 ∂(Π), |∂(Π)| 54 rank of a cell 54 e−, e+ 53 ≡ rank of a map 54 48 reduced admissible word 117 Γ Π 59 reduced band 55 G(i), G(∞) 74 reduced path 56 H 47, 114 g-reduced diagram 75 Ha 116 ( , ) ∗ Z M a -reduced diagram 143 Hka 116 *-reduced word 104 Hra 147 Y-reduced word 67 Hra(∞) 148 T-relator 97 Hkra 151 a-relations 142 Hkra(∞) 164 k-relations 142 K 107 r-relations 142 K(z) 107 ra-relations 142 K 107 S-rule 108 Λ 108 S-rule applicable to an admissible word 109 Λi (0) 123 S-rule locking a sector 109 L(τ, z) 109 S-rule left (right) active for a sector 109 La(τ, z) 118 section of a contour of a diagram 56 M+ 107, 111 + ε-section 96 Mω 107 H-section 98 0(g) 66 sector 108 Ω 107 sector of the type (form) [zz ] 124 (Π1, Γ, Π2) 59 simple in rank i word 74, 97 Π ◦ Θ 152 smooth section 59 R(κ(1)) 116 α-series 63 R(τ, z) 109 side arc of a contiguity subdiagram 59 Ra(τ, z) 118 sides of a trapezium 145 R 108 start and end edges of a band 55 R(z) 110 0-subgroup 65 R(∞) 75 top side of a band 55 Ri 74 trapezium 145 Rkra 164 ψ-trivial word 105 Rra 148 type of a map (diagram) 54 r(Π) 54 van Kampen diagram 54 r(q) 59 van Kampen lemma 54 r(τ, z) 111 vertices in the same phase 77 Si 74 0-word 65 ∼i 74 H-word 96 Θ ∗ Π 151 working rule 107, 111 → 109 τ(z) 108 A 108 top(B) 55 A 107 U(ω) 107 · |A|Y 65 W h 117 A(z) 108 X 108 ¯ α, β, γ, δ, ε, ζ, ι, h, n, α,¯ β,γ¯ 59 Xi 74 NON-AMENABLE FINITELY PRESENTED TORSION-BY-CYCLIC GROUPS 169 z−, z+ 108 Z(M, a) 142 Z(M) 114 Z(M, ra) 151 Z(M, Λ) 114 Z(M, Λ, ra) 151

A. Yu. O. Department of Mathematics Vanderbilt University 1326 Stevenson Center Nashville, TN 37240 USA [email protected] http://www.math.vanderbilt.edu/∼olsh and Department of Higher Algebra, MEHMAT Moscow State University Moscow, Russia [email protected]

M. V. S. Department of Mathematics Vanderbilt University 1326 Stevenson Center Nashville, TN 37240 USA http://www.math.vanderbilt.edu/∼msapir

Manuscrit reçu le 8 février 2001.